Coordinate Geometry Ex 5.4-10th Std Maths-Book Back Question And Answer
Question 1.
Find the slope of the following straight lines.
(i) 5y – 3 = 0
(ii) 7x – 317 = 0
Solution:
(i) 5y – 3 = 0
5y = 3 ⇒ y = 35
Slope = 0
(ii) 7x – 317 = 0 (Comparing with y = mx + c)
7x = 317
Slope is undefined
Question 2.
Find the slope of the line which is
(i) parallel to y = 0.7x – 11
(ii) perpendicular to the line x = -11
Solution:
(i) y = 0.7x – 11
Slope = 0.7 (Comparing with y = mx + c)
(ii) Perpendicular to the line x = – 11
Slope is undefined (Since the line is intersecting the X-axis)
Question 3.
Check whether the given lines are parallel or perpendicular
(i) x3 + y4 + 17 = 0 and 2x3 + y2 + 110 = 0
(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Solution:
(i) x3 + y4 + 17 = 0 ; 2x3 + y2 + 110 = 0
Slope of the line (m1) = −ab
= – 13 ÷ 14 = –13 × 41 = – 43
Slope of the line (m2) = – 23 ÷ 12 = –23 × 21 = – 43
m1 = m2 = – 43
∴ The two lines are parallel.
(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Slope of the line (m1) = −523
Slope of the line (m2) = −23−5 = 235
m1 × m2 = −523 × 235 = -1
∴ The two lines are perpendicular
Question 4.
If the straight lines 12y = -(p + 3)x + 12, 12x – 7y = 16 are perpendicular then find ‘p’
Solution:
Slope of the first line 12y = -(p + 3)x +12
y = −(p+3)x12+1 (Comparing with y = mx + c)
Slope of the second line (m1) = −(p+3)12
Slope of the second line 12x – 7y = 16
(m2) = −ab = −12−7 = 127
Since the two lines are perpendicular
m1 × m2 = -1
−(p+3)12 × 127 = -1 ⇒ −(p+3)7 = -1
-(p + 3) = -7
– p – 3 = -7 ⇒ -p = -7 + 3
-p = -4 ⇒ p = 4
The value of p = 4
Question 5.
Find the equation of a straight line passing through the point P(-5,2) and parallel to the line joining the points Q(3, -2) and R(-5,4).
Solution:
Slope of the line = y2−y1x2−x1
Slope of the line QR = 4+2−5−3 = 6−8 = 3−4 ⇒ – 34
Slope of its parallel = – 34
The given point is p(-5, 2)
Equation of the line is y – y1 = m(x – x1)
y – 2 = – 34 (x + 5)
4y – 8 = -3x – 15
3x + 4y – 8 + 15 = 0
3x + 4y + 7 = 0
The equation of the line is 3x + 4y + 7 = 0
Question 6.
Find the equation of a line passing through (6, -2) and perpendicular to the line joining the points (6, 7) and (2, -3).
Solution:
Let the vertices A (6, 7), B (2, -3), D (6, -2)
Slope of a line = y2−y1x2−x1
Slope of AB = −3−72−6 = −10−4 = 52
Slope of its perpendicular (CD) = – 25
Equation of the line CD is y – y1 = m(x – x1)
y + 2 = –25 (x – 6)
5(y + 2) = -2 (x – 6)
5y + 10 = -2x + 12
2x + 5y + 10 – 12 = 0
2x + 5y – 2 = 0
The equation of the line is 2x + 5y – 2 = 0
Question 7.
A(-3,0) B(10, -2) and C(12,3) are the vertices of ∆ABC. Find the equation of the altitude through A and B.
Solution:
To find the equation of the altitude from A.
The vertices of ∆ABC are A(-3, 0), B(10, -2) and C(12, 3)
Slope of BC = y2−y1x2−x1
= 3+212−10 = 52
Slope of the altitude AD is – 25
Equation of the altitude AD is
y – y1 = m (x – x1)
y – 0 = – 25 (x + 3)
5y = -2x -6
2x + 5y + 6 = 0
Equation of the altitude AD is 2x + 5y + 6 = 0
Equation of the altitude from B
Slope of AC = 3−012+3 = 315 = 15
Slope of the altitude AD is -5
Equation of the altitude BD is y – y1= m (x – x1)
7 + 2 = -5 (x – 10)
y + 2 = -5x + 50
5x + 7 + 2 – 50 = 0 ⇒ 5x + 7 – 48 = 0
Equation of the altitude from B is 5x + y – 48 = 0
Question 8.
Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6, -4).
Solution:
“C” is the mid point of AB also CD ⊥ AB.
Slope of AB = y2−y1x2−x1
= −4−26+4 = −610 = – 35
Slope of the ⊥r AB is 53
Mid point of AB = (x1+x22,y1+y22)
= (−4+62,2−42) = (22,−22) = (1,-1)
Equation of the perpendicular bisector of CD is
y – y1 = m(x – x1)
y + 1 = 53 (x – 1)
5(x – 1) = 3(y + 1)
5x – 5 = 3y + 3
5x – 3y – 5 – 3 = 0
5x – 3y – 8 = 0
Equation of the perpendicular bisector is 5x – 3y – 8 = 0
Question 9.
Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0.
Solution:
Given lines are.
x = 4343 = 1
Substitute the value of x = 1 in (1)
7(1) + 3y = 10 ⇒ 3y = 10 – 7
y = 33 = 1
The point of intersection is (1,1)
Equation of the line parallel to 13x + 5y + 12 = 0 is 13x + 5y + k = 0
This line passes through (1,1)
13 (1) + 5 (1) + k = 0
13 + 5 + k = 0 ⇒ 18 + k = 0
k = -18
∴ The equation of the line is 13x + 5y – 18 = 0
Question 10.
Find the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0.
Solution:
Given lines are.
Substitute the value of x = 167 in (2)
3 × 167 + 2y = 10 ⇒ 2y = 10 – 487
2y = 70−487 ⇒ 2y = 227
y = 222×7 = 117
The point of intersect is (167,117)
Equation of the line perpendicular to 4x – 7y + 13 = 0 is 7x + 4y + k = 0
This line passes through (167,117)
7 (167) + 4 (117) + k = 0 ⇒ 16 + 447 + k = 0
112+447 + k = 0 ⇒ 1567 + k = 0
k = – 1567
Equation of the line is 7x + 4y – 1567 = 0
49x + 28y – 156 = 0
Question 11.
Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y -4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3.
Solution:
The given lines are.
Substitute the value of x = 0 in (1)
3 (0) + y = -2
y = -2
The point of intersection is (0, -2).
The given equation is
Substitute the value of y = 911 in (6)
– x + 2 (911) = 3 ⇒ -x + 1811 = 3
-x = 3 – 1811 = 33−1811 = 1511
x = – 1511
The point of intersection is (-1511,911)
Equation of the line joining the points (0, -2) and (-1511,911) is
31 × (- 11x) = 11 × 15 (y + 2) = 165 (y + 2)
– 341 x = 165 y + 330
– 341 x – 165 y – 330 = 0
341 x + 165 y + 330 = 0
(÷ by 11) ⇒ 31 x + 15 y + 30 = 0
The required equation is 31 x + 15 y + 30 = 0
Question 12.
Find the equation of a straight line through the point of intersection of the lines 8JC + 3j> = 18, 4JC + 5y = 9 and bisecting the line segment joining the points (5, -4) and (-7,6).
Solution:
Given lines are.
8x + 3y = 18 …..(1)
4x + 5y = 9 …..(2)
x = 6328 = 94
Substitute the value of x = 94 in (2)
4 (94) + 5y = 9
9 + 5y = 9 ⇒ 5y = 9 – 9
5y = 0 ⇒ y = 0
The point of intersection is (94,0)
Mid point of the points (5, -4) and (-7, 6)
Equation of the line joining the points (94,0) and (-1,1)
-13y = 4x – 9
-4x – 13y + 9 = 0 ⇒ 4x + 13y – 9 = 0
The equation of the line is 4x + 13y – 9 = 0
