Question

43. Find the equation of the line passing through the point of intersection of the lines
(i) x + y + 1 = 0, 2x – y + 5 = 0 and the point (5,-2)
(ii) 2x - 5y + 1 = 0, x - 3y - 4 = 0 and making equal intercepts on the axes.

iii) Find the equation of the line perpendicular to the line 3x + 4y+6= 0 and making an in
- 4 on the x-axis ​

Answer 1

Answer:

solved

Step-by-step explanation:

(i) x + y + 1 = 0, 2x – y + 5 = 0 and the point (5,-2)

the point of intersection of the lines is (-2,1)

slope = -3/7

y-1/x+2 = -3/7 , y-1 = -3/7(x+2) , 7y-7 = -3x-6

3x+7y-1=0

(ii) 2x - 5y + 1 = 0, x - 3y - 4 = 0 and making equal intercepts on the axes.

the point of intersection of the lines is (-23,-65)

the equation is x+y + 88 = 0

iii) Find the equation of the line perpendicular to the line 3x + 4y+6= 0 and making an in tercept - 4 on the x-axis ​

slope = 4/3

y - 0 ÷ x+4 = 4/3 , 3y = 4x+16

4x-3y+16 = 0