Question

Find the equation of the line passing through the point (2, 3) and perpendicular to the line 3x – 5y = 6.
a) 5x-y-7=0 b) 5x+3y-19=0 c) 7x+5y-1=0 d) 4x+y-6=0
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Answer 1

Answer:

m1 = -a/b

m1 = -3/-5 => 3/5

m1 × m2 = -1

3/5 × m2 = -1

3×m2 =-1×5

3×m2 = -5

m2 = -5/3

y - y1 = m (x - x1)

y - 3 = -5/3 (x - 2)

3(y - 3) = -5 (x - 2)

3y-9 =-5x +10

5x+3y-10-9=0

5x+3y-19=0

Answer 2

The equation is y =  $-\frac{5}{3}$ x + 19/2.

As per the question we need to find the equation of the line who passing through the point (2,3) and perpendicular to the line 2x -5y = 6.

Now

The given line is

2x -5y = 6

we can write the above equation as

5y = 3x - 6

y = $\frac{3}{5}x - \frac{6}{5}$................(1)

The slope of the line (1) is $\frac{3}{5}$ .

A line that perpendicular of the line (1) will have the slope is $-\frac{5}{3}$

Let the equation

y = mx + c

herre m =    $-\frac{5}{3}$

now

y =  $-\frac{5}{3}$x +c .............. (2)

now we find the value of c

Equation 2 passing through the point (2,3).

Therefore

3 =  $-\frac{5}{3}$ * 2 + c

9 + 10 = 2c

c = 19/2

Therefore the equation is y =  $-\frac{5}{3}$ x + 19/2.

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https://brainly.in/question/8461305?

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