Question

Find an equation of the straight line that is perpendicular to the straight line x + 2y = 5 and that passes through the point (3, 7).​

Answer 1

Step-by-step explanation:

refer to the attachment for answer.

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Answer 2

Answer:

Correct option is

D

4x−3y+2=0

The line x+2y=5 and 3x+7y=17 intersect at the point (1,2)

Slope of line 3x+4y=10 is m=−

4

3

The slope of line perpendicular to 3x+4y=10 is m

′

=

m

−1

=

⎝

⎜

⎜

⎛

−

4

3

−1

⎠

⎟

⎟

⎞

=

3

4

So, the equation of line through (1,2) and having slope

3

4

is

(y−2)=

3

4

(x−1)

⟹3(y−2)=4(x−1)

⟹4x−3y+2=0

The answer is option (D)