Question

Find the equation of the straight line perpendicular to the line 2x+3y=0 and passing through the point of intersection of the lines x+3y-1=0 and x-2y+4=0.​

Answer 1

Answer:

x+3y-1=0 and x-2y+4=0

Solve these two equations, you will get x=-2 and y=1

m of this equation 2x+3y=0 is {y=-(2/3)x} -2/3

m of the equation of the line perpendicular to 2x+3y=0 will be 3/2 {m1×m2=-1}

So the required equation of the line will be

y-1=(3/2)(x--2)

y-1=(3/2)(x+2)

2(y-1)=3(x+2)

2y-2=3x+6

3x-2y+8=0 Ans

Answer 2

Answer:

The equation of the straight line is $2y-3x-8=0$

Step-by-step explanation:

Given two lines,

$x+3y-1=0\hspace{1cm} (1)\\\\x-2y+4=0 \hspace{1cm} (2)$

The intersection of these two lines is obtained by solving the above equations simultaneously.

Perform equation (1)-equation (2), to obtain

$\implies 3y-(-2y)-1-4=0\\\\\implies 3y+2y-5=0\\\\\implies 5y=5\\\\\implies y=1$

Substitute $y=1$ in one of the equations to find the value of $x$.

$\implies x+3(1)-1=0\\\\\implies x+3-1=0\\\\\implies x=-2$

So the point of intersection is (-2,1).

Also, the required line is perpendicular to the straight line $2x+3y=0$.

Rewriting the equation,

$\implies 3y=-2x\\\\\implies y=\dfrac{-2}{3}x$

So the gradient of the line is $\frac{-2}{3}$ . Since the product of gradients of perpendicular lines is -1, the gradient of the other line is $\frac{3}{2}$ .

So we need to find the equation of the line with gradient  $\frac{3}{2}$ and passing through the point (-2,1). It is given by,

$\implies y-1=\frac{3}{2}(x-(-2))\\\\\implies y-1= \frac{3}{2}(x+2)\\\\\implies 2(y-1)=3(x+2)\\\\\implies 2y-2=3x+6\\\\\implies 2y-3x-2-6=0\\\\\implies 2y-3x-8=0$