Showing posts with label COORDINATE GEOMETRY. Show all posts
Showing posts with label COORDINATE GEOMETRY. Show all posts

35 Solved Coordinate Geometry Questions for Class 10 Math

35 Solved Coordinate Geometry Questions

1. Find the distance between (2, 3) and (4, 1).
$d = \sqrt{(4-2)^2 + (1-3)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8} = $ 2√2 units
2. Find the distance between (-5, 7) and (-1, 3).
$d = \sqrt{(-1 - (-5))^2 + (3-7)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{32} = $ 4√2 units
3. Distance of point (x, y) from origin (0, 0)?
Distance = $\sqrt{x^2 + y^2}$
4. Find distance between (a, b) and (-a, -b).
$d = \sqrt{(-a-a)^2 + (-b-b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} = $ 2√(a² + b²)
5. Are points (1, 5), (2, 3) and (-2, -11) collinear?
Check if Area of Triangle = 0. $1(3 - (-11)) + 2(-11 - 5) + (-2)(5 - 3) = 14 - 32 - 4 = -22 \neq 0$.
Result: Not Collinear.
6. Find midpoint of (6, 8) and (2, 4).
$M = (\frac{6+2}{2}, \frac{8+4}{2}) = (4, 6)$.
7. Find 'y' if distance between (2, -3) and (10, y) is 10.
$10^2 = (10-2)^2 + (y+3)^2 \Rightarrow 100 = 64 + (y+3)^2 \Rightarrow 36 = (y+3)^2$.
$y+3 = \pm 6 \Rightarrow y = 3$ or $y = -9$.
8. Relation between x and y such that (x, y) is equidistant from (7, 1) and (3, 5)?
$(x-7)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2$. Simplifying: x - y = 2.
9. Point on x-axis equidistant from (2, -5) and (-2, 9).
Point is (x, 0). $(x-2)^2 + (0+5)^2 = (x+2)^2 + (0-9)^2 \Rightarrow -8x = 56 \Rightarrow x = -7$.
Point: (-7, 0).
10. Find the centroid of triangle with vertices (1,1), (0,0), and (2,2).
$G = (\frac{1+0+2}{3}, \frac{1+0+2}{3}) = (1, 1)$.
11. Find coordinates dividing (-1, 7) and (4, -3) in ratio 2:3.
$x = \frac{2(4)+3(-1)}{5} = 1, y = \frac{2(-3)+3(7)}{5} = 3$. Point: (1, 3).
12. Coordinates of points of trisection of (4, -1) and (-2, -3).
Ratio 1:2 $\Rightarrow$ (2, -5/3); Ratio 2:1 $\Rightarrow$ (0, -7/3).
13. Find ratio in which y-axis divides (-4, 5) and (3, -7).
On y-axis, x=0. $0 = \frac{k(3) + 1(-4)}{k+1} \Rightarrow 3k = 4 \Rightarrow k = 4:3$.
14. Ratio in which (-3, 10) and (6, -8) is divided by (-1, 6).
$-1 = \frac{6k - 3}{k+1} \Rightarrow -k-1 = 6k-3 \Rightarrow 7k=2 \Rightarrow$ 2:7.
15. Area of triangle with vertices (2, 3), (-1, 0), (2, -4).
$\frac{1}{2}|2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)| = \frac{1}{2}|8 + 7 + 6| = $ 10.5 sq. units.
16. Find 'k' if (2, 3), (4, k), (6, -3) are collinear.
Area = 0 $\Rightarrow 2(k+3) + 4(-3-3) + 6(3-k) = 0 \Rightarrow -4k = 0 \Rightarrow k = 0$.
17. Perimeter of triangle with vertices (0,4), (0,0), (3,0).
Sides: 4, 3, and $\sqrt{3^2+4^2}=5$. Perimeter = $4+3+5 = 12$ units.
18. Find fourth vertex of parallelogram: (1,2), (4,3), (6,6).
Midpoints of diagonals are same. $(1+6)/2 = (4+x)/2 \Rightarrow x=3$. $(2+6)/2 = (3+y)/2 \Rightarrow y=5$. Vertex: (3, 5).
19. Coordinate of point A where AB is diameter, center is (2, -3), B is (1, 4).
$2 = (x+1)/2 \Rightarrow x=3$. $-3 = (y+4)/2 \Rightarrow y=-10$. A is (3, -10).
20. Distance between $(a\cos\theta, 0)$ and $(0, a\sin\theta)$.
$\sqrt{(0-a\cos\theta)^2 + (a\sin\theta-0)^2} = \sqrt{a^2(\cos^2\theta + \sin^2\theta)} = \sqrt{a^2} = a$.
21. If P(x,y) is equidistant from A(5,1) and B(-1,5), find relation.
$(x-5)^2 + (y-1)^2 = (x+1)^2 + (y-5)^2 \Rightarrow 3x = 2y$.
22. Find area of rhombus if vertices are (3,0), (4,5), (-1,4), (-2,-1).
$\frac{1}{2} \times d_1 \times d_2$. $d_1 = \sqrt{32}, d_2 = \sqrt{72}$. Area = 24 sq. units.
23. Point P divides AB in ratio 1:3. A(2,1), B(7,6). Find P.
$x = \frac{1(7)+3(2)}{4} = 3.25, y = \frac{1(6)+3(1)}{4} = 2.25$.
24. Find distance of (4, -3) from x-axis.
Distance from x-axis = $|y| = |-3| = 3$ units.
25. Find distance of (4, -3) from y-axis.
Distance from y-axis = $|x| = 4$ units.
26. If points (0,0), (3,√3) and (3,p) form equilateral triangle, find p.
Distance OP = PQ. $p = -\sqrt{3}$.
27. Midpoint of line joining (2a, 4) and (-2, 3b) is (1, 2a+1). Find a, b.
$(2a-2)/2 = 1 \Rightarrow a=2$. $(4+3b)/2 = 2(2)+1 \Rightarrow b=2$.
28. Find area of triangle with vertices (a, b+c), (b, c+a), (c, a+b).
$Area = 0$ (Points are collinear).
29. Find distance between $(L, M)$ and $(L+a, M+b)$.
$d = \sqrt{(L+a-L)^2 + (M+b-M)^2} = \sqrt{a^2+b^2}$.
30. Coordinate of point on y-axis equidistant from (6,5) and (-4,3).
Point (0, y). $6^2 + (y-5)^2 = (-4)^2 + (y-3)^2 \Rightarrow y = 9$. Point: (0, 9).
31. Ratio in which line segment joining (1,-5) and (-4,5) is divided by x-axis.
$y=0 \Rightarrow 0 = \frac{k(5) + 1(-5)}{k+1} \Rightarrow k=1$. Ratio: 1:1.
32. Centroid of triangle with vertices (a,b), (b,c), (c,a) is origin. Find a+b+c.
$(a+b+c)/3 = 0 \Rightarrow a+b+c = 0$.
33. Distance between $(\cos\theta, \sin\theta)$ and $(\sin\theta, \cos\theta)$.
$\sqrt{(\sin\theta-\cos\theta)^2 + (\cos\theta-\sin\theta)^2} = \sqrt{2(\sin\theta-\cos\theta)^2} = \sqrt{2}|\sin\theta-\cos\theta|$.
34. Find 'p' if (p, 2) is midpoint of (5, 3) and (1, 1).
$p = (5+1)/2 = 3$.
35. Type of triangle formed by (1,1), (-1,-1), (√3, -√3).
$AB = \sqrt{8}, BC = \sqrt{8}, AC = \sqrt{8}$. Result: Equilateral Triangle.
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29 Coordinate Geometry. Complete Coordinate Geometry: 10 Solved Examples & 50 Practice Questions (Class 10)

Coordinate Geometry: Complete Guide for Class 10

Master Coordinate Geometry with these 10 fully solved examples covering Distance Formula, Section Formula, and Area of Triangles. Afterward, test your skills with 50 practice questions provided with an answer key.

Part 1: 10 Fully Solved Important Questions

Q1 - Distance Formula
Find the distance between the points $A(2, 3)$ and $B(4, 1)$.
Solution: Let $A(x_1, y_1) = (2, 3)$ and $B(x_2, y_2) = (4, 1)$.
Using the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$d = \sqrt{(4 - 2)^2 + (1 - 3)^2}$$ $$d = \sqrt{(2)^2 + (-2)^2}$$ $$d = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ units.}$$
Q2 - Midpoint Formula
Find the coordinates of the midpoint of the line segment joining $P(-5, 7)$ and $Q(-1, 3)$.
Solution: Using the midpoint formula $M(x, y) = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$: $$x = \frac{-5 + (-1)}{2} = \frac{-6}{2} = -3$$ $$y = \frac{7 + 3}{2} = \frac{10}{2} = 5$$ The midpoint is $(-3, 5)$.
Q3 - Section Formula
Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ in the ratio $2:3$.
Solution: Let the points be $A(-1, 7)$ and $B(4, -3)$. Ratio $m_1:m_2 = 2:3$.
Using Section Formula: $$x = \frac{m_1x_2 + m_2x_1}{m_1+m_2}, \quad y = \frac{m_1y_2 + m_2y_1}{m_1+m_2}$$ $$x = \frac{2(4) + 3(-1)}{2+3} = \frac{8 - 3}{5} = \frac{5}{5} = 1$$ $$y = \frac{2(-3) + 3(7)}{2+3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3$$ The required point is $(1, 3)$.
Q4 - Value of k
Find the value of $k$ if the points $A(2, 3)$, $B(4, k)$, and $C(6, -3)$ are collinear.
Solution: For collinear points, the area of the triangle formed by them is 0. $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$$ $$2(k - (-3)) + 4(-3 - 3) + 6(3 - k) = 0$$ $$2(k + 3) + 4(-6) + 18 - 6k = 0$$ $$2k + 6 - 24 + 18 - 6k = 0$$ $$-4k = 0 \Rightarrow k = 0$$
Q5 - Equidistant Points
Find a point on the y-axis which is equidistant from the points $A(6, 5)$ and $B(-4, 3)$.
Solution: Let the point on y-axis be $P(0, y)$.
Given $PA = PB$, so $PA^2 = PB^2$. $$(6 - 0)^2 + (5 - y)^2 = (-4 - 0)^2 + (3 - y)^2$$ $$36 + 25 + y^2 - 10y = 16 + 9 + y^2 - 6y$$ $$61 - 10y = 25 - 6y$$ $$36 = 4y \Rightarrow y = 9$$ The point is $(0, 9)$.
Q6 - Ratio Finding
In what ratio does the point $(-4, 6)$ divide the line segment joining the points $A(-6, 10)$ and $B(3, -8)$?
Solution: Let the ratio be $k:1$. Using the Section Formula for the x-coordinate: $$-4 = \frac{k(3) + 1(-6)}{k+1}$$ $$-4(k+1) = 3k - 6$$ $$-4k - 4 = 3k - 6$$ $$2 = 7k \Rightarrow k = \frac{2}{7}$$ Therefore, the ratio is $2:7$.
Q7 - Parallelogram Vertex
If $(1, 2)$, $(4, y)$, $(x, 6)$, and $(3, 5)$ are vertices of a parallelogram taken in order, find $x$ and $y$.
Solution: Diagonals of a parallelogram bisect each other. Midpoint of AC = Midpoint of BD. $$\left(\frac{1+x}{2}, \frac{2+6}{2}\right) = \left(\frac{4+3}{2}, \frac{y+5}{2}\right)$$ $$\frac{1+x}{2} = \frac{7}{2} \Rightarrow 1+x = 7 \Rightarrow x = 6$$ $$\frac{8}{2} = \frac{y+5}{2} \Rightarrow 8 = y+5 \Rightarrow y = 3$$ So, $x=6, y=3$.
Q8 - Centroid of Triangle
Find the centroid of the triangle formed by the vertices $(3, -5)$, $(-7, 4)$, and $(10, -2)$.
Solution: Centroid $G(x,y) = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$. $$x = \frac{3 + (-7) + 10}{3} = \frac{6}{3} = 2$$ $$y = \frac{-5 + 4 + (-2)}{3} = \frac{-3}{3} = -1$$ Centroid is $(2, -1)$.
Q9 - Circle Diameter
Find the coordinates of a point A, where AB is the diameter of a circle whose center is $(2, -3)$ and B is $(1, 4)$.
Solution: Let $A = (x, y)$. The center $C(2, -3)$ is the midpoint of $AB$. $$\frac{x+1}{2} = 2 \Rightarrow x+1=4 \Rightarrow x=3$$ $$\frac{y+4}{2} = -3 \Rightarrow y+4=-6 \Rightarrow y=-10$$ Point A is $(3, -10)$.
Q10 - Area of Triangle
Find the area of a triangle whose vertices are $(1, -1)$, $(-4, 6)$, and $(-3, -5)$.
Solution: Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ $$= \frac{1}{2} |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|$$ $$= \frac{1}{2} |1(11) - 4(-4) - 3(-7)|$$ $$= \frac{1}{2} |11 + 16 + 21| = \frac{1}{2} |48| = 24 \text{ sq units.}$$

Part 2: 50 Practice Questions

  • Find the distance between points (0, 0) and (36, 15).
  • Find the distance between (a, b) and (-a, -b).
  • Calculate the distance of point P(6, -6) from the origin.
  • Find x if the distance between (x, 7) and (1, 15) is 10 units.
  • Find the midpoint of the line segment joining (3, 4) and (5, 2).
  • The midpoint of line segment joining (2a, 4) and (-2, 2b) is (1, 2a+1). Find a and b.
  • Find the centroid of the triangle with vertices (1, 4), (-1, -1), and (3, -2).
  • Find the coordinates of the point dividing the line joining (-1, 3) and (4, -7) in ratio 3:4.
  • Determine the ratio in which the line segment joining (1, -5) and (-4, 5) is divided by the x-axis.
  • Find the coordinates of the point of trisection of the line segment joining (2, -2) and (-7, 4).
  • Find the area of the triangle formed by vertices (2, 3), (-1, 0), (2, -4).
  • Find the value of k if points (7, -2), (5, 1), and (3, k) are collinear.
  • If the distance between (4, p) and (1, 0) is 5, find p.
  • Check if the points (5, -2), (6, 4), and (7, -2) form an isosceles triangle.
  • Find a point on the x-axis which is equidistant from (2, -5) and (-2, 9).
  • Find the perimeter of the triangle with vertices (0, 0), (3, 0), and (0, 4).
  • Find the fourth vertex of the rectangle with three vertices (0,0), (2,0), and (0,3).
  • The coordinates of one end of a diameter of a circle are (2, 3) and the center is (-2, 5). Find the other end.
  • Find the value of y for which the distance between P(2, -3) and Q(10, y) is 10 units.
  • Show that points (1, 7), (4, 2), (-1, -1) are vertices of a square (Distance check).
  • Find the ratio in which the y-axis divides the line segment joining (5, -6) and (-1, -4).
  • If (1, 2), (4, y), (x, 6), and (3, 5) are vertices of a parallelogram, find x+y.
  • Find the distance of point (2, 3) from the x-axis.
  • Find the distance of point (-5, 4) from the y-axis.
  • What is the distance between the points ($a \cos \theta, 0$) and ($0, a \sin \theta$)?
  • Find the centroid of a triangle with vertices (3, -7), (-8, 6), and (5, 10).
  • If the origin is the centroid of the triangle with vertices (x, 1), (y, -2), (2, 3), find x and y.
  • The line segment joining (2, -3) and (5, 6) is divided by the x-axis in what ratio?
  • Find the coordinates of a point on the x-axis which is equidistant from (5, 4) and (-2, 3).
  • The area of a triangle with vertices (a, 0), (0, b), and (1, 1) is collinear. Find the relation between a and b.
  • Name the type of triangle formed by (3, 2), (-2, -3), (2, 3).
  • If P(9a - 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3:1, find a and b.
  • Find the distance between A(2a, 6a) and B(2a + \sqrt{3}a, 5a).
  • If A(-2, 1), B(a, 0), C(4, b), and D(1, 2) are vertices of a parallelogram, find a and b.
  • Find the coordinates of the point which is equidistant from the three vertices of $\Delta$ AOB where A=(0,2y), O=(0,0), B=(2x,0).
  • Find the area of the triangle formed by (0, 0), (4, 0), and (0, 3).
  • Points A(4, 3), B(6, 4), C(5, -6) and D(-3, 5) are vertices of a parallelogram? (True/False).
  • Find the perpendicular distance of A(5, 12) from the origin.
  • Find the value of k for which A(-5, 1), B(1, k), and C(4, -2) are collinear.
  • The midpoint of (3p, 4) and (-2, 2q) is (2, 6). Find p + q.
  • Find the coordinates of the circumcenter of the triangle formed by (0, 0), (4, 0), and (0, 4).
  • In what ratio does the point P(2, -5) divide the line joining A(-3, 5) and B(4, -9)?
  • The distance between points (5, 3) and (x, -1) is 5. Find x.
  • Find the area of the quadrilateral ABCD with vertices A(-5, 7), B(-4, -5), C(-1, -6), and D(4, 5).
  • Determine if the points (1, 5), (2, 3), and (-2, -11) are collinear.
  • Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) such that AP = 2/5 AB.
  • Find the relation between x and y such that the point (x, y) is equidistant from (7, 1) and (3, 5).
  • Find the coordinates of the point of intersection of the medians of a triangle with vertices (-1, 0), (5, -2), and (8, 2).
  • If the area of a triangle formed by (x, 2x), (-2, 6), and (3, 1) is 5 sq units, find x.
  • Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).

Answer Key

Q.NoAnswer Q.NoAnswer Q.NoAnswer Q.NoAnswer Q.NoAnswer
1391110.5 sq units215:131scalene41p=2, q=4, sum=6
2$2\sqrt{a^2+b^2}$12422932a=1, b=-342(2, 2)
3$6\sqrt{2}$13$\pm 4$233 units332a432:5 approx (check coords)
47 or -514Yes245 units34a=1, b=1442 or 8
5(4, 3)15(-7, 0)25a35(x, y)4572 sq units
6a=2, b=21612 units26(0, 3)366 sq units46Yes
7(1, 0)17(2, 3)27x=-2, y=-137False47(3, 4)
8(1.14, -1.28)18(-6, 7)281:2381348x - y = 2
91:1193 or -929(2, 0) approx39-149(4, 0)
10(-1, 0) & (-4, 2)20Proof301/a + 1/b = 140650(-7, 0)
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8 Essential Coordinate Geometry Problems with Solutions and Formulas

Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. This integration of algebra and geometry allows us to solve geometric problems using algebraic equations. Below are the essential formulas and 8 key problems with detailed solutions to help you master the basics.

Key Formulas Cheat Sheet

Distance Formula $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Section Formula (Internal Division) $$P(x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right)$$
Midpoint Formula $$M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$
Area of a Triangle $$Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Solved Problems

Problem 1: Distance Between Two Points

Find the distance between the points $A(2, 3)$ and $B(5, 7)$.

Solution:

Let $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (5, 7)$.

Using the distance formula:

$$d = \sqrt{(5 - 2)^2 + (7 - 3)^2}$$ $$d = \sqrt{(3)^2 + (4)^2}$$ $$d = \sqrt{9 + 16}$$ $$d = \sqrt{25}$$
Distance = 5 units
Problem 2: Midpoint Calculation

Find the coordinates of the midpoint of the line segment joining the points $P(-4, 2)$ and $Q(8, 6)$.

Solution:

Using the midpoint formula $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$:

$$x = \frac{-4 + 8}{2} = \frac{4}{2} = 2$$ $$y = \frac{2 + 6}{2} = \frac{8}{2} = 4$$
Midpoint M = (2, 4)
Problem 3: Section Formula (Ratios)

Find the coordinates of the point which divides the line segment joining the points $(4, -3)$ and $(8, 5)$ in the ratio $3:1$ internally.

Solution:

Here, $x_1=4, y_1=-3, x_2=8, y_2=5$, and ratio $m_1:m_2 = 3:1$.

Using the Section Formula:

$$x = \frac{3(8) + 1(4)}{3 + 1} = \frac{24 + 4}{4} = \frac{28}{4} = 7$$ $$y = \frac{3(5) + 1(-3)}{3 + 1} = \frac{15 - 3}{4} = \frac{12}{4} = 3$$
Coordinates = (7, 3)
Problem 4: Determining Collinearity

Determine if the points $A(1, 5)$, $B(2, 3)$, and $C(-2, 11)$ are collinear.

Solution:

Points are collinear if the area of the triangle formed by them is zero.

$$Area = \frac{1}{2} |1(3 - 11) + 2(11 - 5) + (-2)(5 - 3)|$$ $$Area = \frac{1}{2} |1(-8) + 2(6) - 2(2)|$$ $$Area = \frac{1}{2} |-8 + 12 - 4|$$ $$Area = \frac{1}{2} |0| = 0$$
Since Area = 0, the points are collinear.
Problem 5: Finding a Missing Coordinate (Equidistant)

Find the value of $x$ if the point $(x, 2)$ is equidistant from $(8, -2)$ and $(2, -2)$.

Solution:

Let $P(x, 2)$, $A(8, -2)$, and $B(2, -2)$. We are given $PA = PB$, so $PA^2 = PB^2$.

$$(x - 8)^2 + (2 - (-2))^2 = (x - 2)^2 + (2 - (-2))^2$$ $$(x - 8)^2 + 4^2 = (x - 2)^2 + 4^2$$

Subtract $16$ ($4^2$) from both sides:

$$(x - 8)^2 = (x - 2)^2$$ $$x^2 - 16x + 64 = x^2 - 4x + 4$$

Cancel $x^2$ and solve for $x$:

$$-16x + 4x = 4 - 64$$ $$-12x = -60$$ $$x = 5$$
x = 5
Problem 6: Centroid of a Triangle

Find the centroid of a triangle whose vertices are $(3, -5)$, $(-7, 4)$, and $(10, -2)$.

Solution:

The formula for the centroid is $G(x, y) = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)$.

$$x = \frac{3 + (-7) + 10}{3} = \frac{6}{3} = 2$$ $$y = \frac{-5 + 4 + (-2)}{3} = \frac{-3}{3} = -1$$
Centroid G = (2, -1)
Problem 7: Area of a Triangle

Calculate the area of the triangle formed by the vertices $A(2, 3)$, $B(-1, 0)$, and $C(2, -4)$.

Solution:

Using the area formula:

$$Area = \frac{1}{2} |2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)|$$ $$Area = \frac{1}{2} |2(4) - 1(-7) + 2(3)|$$ $$Area = \frac{1}{2} |8 + 7 + 6|$$ $$Area = \frac{1}{2} |21|$$
Area = 10.5 square units
Problem 8: Ratio of Division by Axis

Find the ratio in which the y-axis divides the line segment joining the points $(5, -6)$ and $(-1, -4)$.

Solution:

Any point on the y-axis has coordinates $(0, y)$. Let the ratio be $k:1$.

Using the section formula for the x-coordinate (which we know is 0):

$$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}$$ $$0 = \frac{k(-1) + 1(5)}{k + 1}$$

Cross multiply:

$$0 = -k + 5$$ $$k = 5$$
The ratio is 5:1
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COORDINATE GEOMETRY

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