Question

Which of these is a point-slope equation of the line that is perpendicular to

y-25=2(x-10) and passes through (-3, 7)?
O Ay+ 7 = -(x-3)
OB. y-7=-(x+3)
O C. y+ 7 = 2(x-3)
OD. y-7= -2(x+3)

Answer 1

`y -7= (-1)/(2) (x+3)`

Explanation:

The given equation of the line is ,

`\implies y - 25 = 2(x-10) \\`

Simplify the RHS by opening the brackets,

`\implies y - 25 = 2x - 20\\`

`\implies y = 2x + 25 - 20 \\`

`\implies y = 2x + 5 \\`

On comparing it with the the slope intercept form of the line, namely y = mx + c , where m is the slope, we have;

`\implies m = 2\\`

Again as we know that the product of slopes of two perpendicular lines is -1 , Hence,

`\implies m\cdot m_(\perp) = 2 \\`

`\implies m_(\perp) =(-1)/(2) \\`

Now the point given to us is (-3,7) , the point slope form of the line is ,

`\implies y - y_1 = m(x-x_1) \\`

where ,

• m is the slope.
• (x1,y1) is the point through which the line passes.

On substituting the respective values, we have;

`\implies y - 7 = (-1)/(2) \{ x -(-3)\}\\`

Simplify,

`\implies \underline{\underline{ y - 7 =(-1)/(2)(x+3)}}\\`

This is the required answer in point slope form .

Answer 2

`y-7=-(1)/(2)(x+3)`

Explanation:

The point-slope formula is:

`y-y_1=m(x-x_1)`

where:

• m is the slope.
• (x₁, y₁) is a point on the line.

The equation of the original line in point-slope form is:

`y-25=2(x-10)`

Therefore, the slope of the line is m = 2.

As the slope of a perpendicular line is the negative reciprocal of the slope of the original line, the slope of the perpendicular line is m = -1/2.

To find the point-slope equation of the line that is perpendicular to the given line and passes through point (-3, 7), substitute m = -1/2 and point (-3, 7) into the point-slope formula:

`\implies y-7=-(1)/(2)(x-(-3)`

`\implies y-7=-(1)/(2)(x+3)`

Therefore, the point-slope equation of the line that is perpendicular to y - 25 = 2(x - 10) and passes through (-3, 7) is:

`\boxed{y-7=-(1)/(2)(x+3)}`