### Practice Set 5.3 Co-ordinate Geometry Class 10th Mathematics Part 2 MHB Solution

Practice Set 5.3

1. Angles made by the line with the positive direction of X-axis are given. Find the slope…
2. Find the slopes of the lines passing through the given points. (1) A (2, 3), B (4, 7)…
3. Determine whether the following points are collinear. (1) A(-1, -1), B(0, 1), C(1, 3)…
4. If A (1, -1),B (0, 4),C (-5, 3) are vertices of a triangle then find the slope of each…
5. Show that A (-4, -7),B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a…
6. Find k, if R(1, -1), S (-2, k) and slope of line RS is -2.
7. Find k, if B(k, -5), C (1, 2) and slope of the line is 7.
8. Find k, if PQ || RS and P(2, 4), Q (3, 6), R(3, 1), S(5, k) .

###### Practice Set 5.3
Question 1.

Angles made by the line with the positive direction of X-axis are given. Find the slope of these lines.

(1) 45°

(2) 60°

(3) 90°

slope is given as the tangent of the angle formed with the positive direction of x-axis

1. tan 45° = 1

2. tan60° = 3. tan 90° = cannot be determined.

Question 2.

Find the slopes of the lines passing through the given points.
(1) A (2, 3), B (4, 7)

(2) P (-3, 1), Q (5, -2)

(3) C (5, -2), D (7, 3)

(4) L (-2, -3), M (-6, -8)

(5) E(-4, -2), F (6, 3)

(6) T (0, -3), S (0, 4)

Slope m of a line passing through two points A(a,b) and B(c,d) ig given by 1. A (2, 3), B (4, 7) 2. P (-3, 1), Q (5, -2) 3. C (5, -2), D (7, 3) 4. L (-2, -3), M (-6, -8) 5. E(-4, -2), F (6, 3) 6. T (0, -3), S (0, 4) As denominator is 0,

So, slope cannot be determined.

Question 3.

Determine whether the following points are collinear.
(1) A(-1, -1), B(0, 1), C(1, 3)

(2) D(-2, -3), E(1, 0), F(2, 1)

(3) L(2, 5), M(3, 3), N(5, 1)

(4) P(2, -5), Q(1, -3), R(-2, 3)

(5) R(1, -4), S(-2, 2), T(-3, 4)

(6) A(-4, 4), K(-2, 5/ 2 ), N(4, -2)

Three points are said to be collinear if they all lie in a straight line.
If Three points (x1, y1), (x2y2), (x3y3) are collinear then no triangle can be formed using three points and so the area formed by the triangle by the three points is zero.

Area of Triangle = 1/2 [x(y− y3) + x(y− y1) + x(y− y2)] ...(1)

1. For triangle, A(-1, -1), B(0, 1), C(1, 3)
Area = = 1/2 [ -1(-2) + 0(3+1) + 1(-1-1)]

= 1/2 [2 + 0 - 2]

= 1/2 [2-2]
= 0
Hence the points are collinear

2. For triangle, D(-2, -3), E(1, 0), F(2, 1)
Using 1,
Area = 1/2 [-2(0-1) + 1(1-(-3)) + 2(-3-0)]
= 1/2 [-2(-1) + 1(1+3) + 2(-3)]
= 1/2 [ 2 + 4 -6 ]
= 1/2 [ 6 - 6 ]
= 1/2 (0)
= 0
Hence the points are collinear.

3. For triangle, L(2, 5), M(3, 3), N(5, 1)
Using 1,

Area = 1/2 [2(3-1) + 3(1-5) + 5(5-3)]
= 1/2 [ 2(2) + 3(-4) + 5(2)]
= 1/2 [ 4 - 12 + 10 ]
= 1/2 [ 14 - 12 ]

= 1/2 (2)
= 1

Hence the points are not collinear.

4. For triangle, P(2, -5), Q(1, -3), R(-2, 3)
Using 1,
Area = = 0
Area = 1/2 [ 2(-3-3) + 1(3-(-5)) + (-2)(-5-(-3))]
= 1/2 [ 2(-6) + 1(3+5) - 2 (-5+3)]
= 1/2 [ -12 + 8 -2(-2)]
= 1/2 [-12 + 8 +4]
= 1/2 [ -12 + 12]
= 1/2 (0)

= 0

Hence the points are collinear.

5. For triangle, R(1, -4), S(-2, 2), T(-3, 4)
Area = Hence the points are collinear.

6. For triangle, A(-4, 4), K(-2, 5/ 2 ), N(4, -2)
Area = Hence the points are collinear.

Question 4.

If A (1, -1),B (0, 4),C (-5, 3) are vertices of a triangle then find the slope of each side

Slope m of a line passing through two points A(a,b) and B(c,d) is given by Slope of AB = Slope of BC = Slope of AC = Question 5.

Show that A (-4, -7),B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram.

In a parallelogram, opposite sides are equal and parallel.

According to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by .....(1)

Slope m of a line passing through two points A(a,b) and B(c,d) ig given by In the question,

AB = = BC = = CD = = AD = = Slope of AB = Slope of BC = Slope of CD = Slope of AD = As AB = DC and BC = AD

And Slope AB = Slope CD

Hence the given points form a parallelogram.

Question 6.

Find k, if R(1, -1), S (-2, k) and slope of line RS is -2.

Slope m of a line passing through two points A(a,b) and B(c,d) ig given by In the given question Simplifying

6 = k + 1

K = 5

Question 7.

Find k, if B(k, -5), C (1, 2) and slope of the line is 7.

Slope m of a line passing through two points A(a,b) and B(c,d) ig given by In the given question Simplifying

7-7k = 7

k = 0

Question 8.

Find k, if PQ || RS and P(2, 4), Q (3, 6), R(3, 1), S(5, k) .

two lines are said to be parallel if their slopes are equal

If PQ||RS then their slopes must be equal

Slope of PQ = Slope pf RS = As their slopes are equal, we get Simplifying

k-1 = 4

k = 5

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