Question

Find the equation of line passing through (-2,3) and which is perpendicular to the line 3x+4y =5​

Answer 1

Answer:

Given point is : (2,3)

The given equation is : 3x + 4y - 5 = 0.

So, first we find the slope of the equation,

3x + 4y - 5 = 0

4y = 5 - 3x

So, y = 5/4 - 3x/4

The Slope is : -3/4

By the condition of perpendicularity,

M1 × M2 = -1

-3/4 × M2 = -1

M2 = 4/3

The Slope of perpendicular condition ( M2) is : 4/3.

Given point ( X1 , Y1 ) = (2,3).

On finding the equation ,

So, we know that : Y - Y1 = M2 ( X - X1 )

y - 3 = 4/3 ( x - 2 )

3y - 9 = 4x - 8

-9 +8 = 4x - 3y

4x - 3y = -1

4x - 3y + 1 = 0.

The resultant equation is : 4x - 3y + 1 =0.

Step-by-step explanation:

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Answer 2

$\large\bf{\underline\orange{Answer \:↝}}$

Que :- Find the equation of line passing through (-2,3) and which is perpendicular to the line 3x+4y =5.

Ans :- The slope of the given line 3x + 4y = 5 is found by putting this in slope-intercept form

Move the x to the right side of the equation by subtracting 3x from each side

4y = -3x + 5

So the same equation in that form is y = -(3/4)x + 5/4

This tells us the slope is -3/4

Now

We use the same formula but we use the negative reciprocal of -3/4 for the slope to find a perpendicular line

y - 3 = (4/3) (x + 2)

y - 3 = 4x/3 + 8/3

Multiply through by 3

3y - 9 = 4x + 8

3y = 4x + 17

in slope intercepet form: y = 4x/3 + 17/3 in standard for: 4x - 3y = -17