Question #141773
find the equation of the straight line that passes through the intersection point between the lines x + y = 3 and 2x - 3y = 16 and perpendicular to the straight line y = 4x - 3
1
Expert's answer
2020-11-02T20:42:10-0500

x + y = 3 in the form y=mx+c

will be y=x+3y=-x+3 ........(equation 1)


2x - 3y = 16 in the form y=mx+c

will be y=23x163y=\frac{2}{3}x - \frac{16}{3}........(equation 2)

At point of intersection (equation 1) is equal to (equation 2)


x+3=23x163-x+3=\frac{2}{3}x - \frac{16}{3}


3+165=23x+x3+\frac{16}{5} = \frac{2}{3}x+x


253=53x\frac{25}{3} =\frac{5}{3}x


Hence. x=5x = 5

From (equation 1) y=x+3y=-x+3

Therefore

y=5+3y = -5 +3

y=2y = -2

Hence point of intersection is (5,-2)



For perpendicular lines the product of their gradient is -1

For line y=4x3y=4x-3 the gradient is 44

Let the gradient of the other line be M2


M2 x 4 = -1

M2 = -1/4


y2x5=14\frac{y--2}{x-5} = \frac{-1}{4}


4y+8=x+54y+8=-x+5

4y=x34y = -x-3


The equation of the line will be,

y=14x34y= -\frac{1}{4}x -\frac{3}{4}















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