M.L Aggarwal book Important Question of class 10 chapter Based on Equation of a Straight Line for ICSE BOARD.
You have given an equation of a line.
So , Find the the slope of the line and the equation of a line perpendicular to the given line and passing through the intersection of the given lines.
This is the Question Number 22, Exercise 12.2 of M.L Aggarwal.
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The equation of a line is 3x + 4y – 7 = 0. Find (i) the slope of the line. (ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.
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Given line equation: 3x + 4y – 7 = 0
(i) Slope of the line is given by,
4y = -3x + 7
y = (-3/4) x + 7
Hence, slope (m1) = -3/4
(ii) Let the slope of the perpendicular to the given line be m2
Then, m1 x m2 = -1
(-3/4) x m2 = -1
m2 = 4/3
Now, to find the point of intersection of
x – y + 2 = 0 … (i)
3x + y – 10 = 0 … (ii)
On adding (i) and (ii), we get
4x – 8 = 0
4x = 8
x = 8/4 = 2
Putting x = 2 in (i), we get
2 – y + 2 = 0
y = 4
Hence, the point of intersection of the lines is (2, 4)
The equation of the line having slope m2 and passing through (2, 4) will be
y – 4 = (4/3) (x – 2)
3y – 12 = 4x – 8
4x – 3y + 4 = 0
Thus, the required line equation is 4x – 3y + 4 = 0.