Write the standard form of the equation of the line described. through: (2,4), perp to y=-2/7x-5

The first line equation: y = -2/7x - 5; m1 = -7/2
The second line passes through (2,4) and is perpendicular to the first line.

Therefore,
The gradient of first line = -1/m1 = -1/(-2/7) = 7/2

Then, the equation of the  second line is;
y = 7/2x+b

Substituting for the coordinates given;
4 = 7/2(2)+b
4 = 7+b
b = 4-7 = -3

Therefore, the equation of the second line is;
y = 7/2x - 3

In standard form, the equation this equation can be written as:
y(2) = 7x - 3(2)
2y = 7x - 6
7x -2y - 6 = 0 > standard form.

Answer: c. x y2 04 66 16