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AfilCa [17]
3 years ago
7

Passing through (2, - 2) and perpendicular to the line whose equation is y= 5x+2

Mathematics
1 answer:
jonny [76]3 years ago
7 0

Step-by-step explanation:

Hey there!!!

Here,

Given, A line passes through point (2,-2) and is perpendicular to the y= 5x+2.

The equation of a straight line passing through point is,

(y - y1) = m1(x - x1)

Now, put all values.

(y  + 2) = m1(x - 2)

It is the 1st equation.

Another equation is;

y = 5x +2........(2nd equation).

Now, Comparing it with y = mx + c, we get;

m2=5

As per the condition of perpendicular lines,

m1×m2= -1

m1 × 5 = -1

Therefore, m2= -1/5.

Keeping the value of m1 in 1st equation.

(y + 2) =  \frac{ - 1}{5} (x - 2)

Simplify them.

5(y + 2) =  - x + 2

5y + 10 =  - x + 2

x + 5y + 8 = 0

Therefore the required equation is x+5y+8= 0.

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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