In the xy-plane, line l_1 passes through the origin and has a positive slope m, where m > … — SAT

In the $xy$ -plane, line $l_1$ passes through the origin and has a positive slope $m$ , where $m > 1$ . Line $l_2$ is perpendicular to line $l_1$ and has a $y$ -intercept of $(0, 10)$ . The two lines intersect at point $P$ . If the distance from point $P$ to the $y$ -axis is $4$ , what is the value of $m$ ?

Answer: 2

Answer

The equation of line

l_1

y = mx

and the equation of line

l_2

y = -\frac{1}{m}x + 10

. Setting these equal gives the

x

-coordinate of their intersection as

x = \frac{10m}{m^2 + 1}

. Since the distance from the intersection point to the

y

-axis is

4

, we have

\frac{10m}{m^2 + 1} = 4

. Solving this quadratic equation yields

m = 2

m = \frac{1}{2}

. Given that

m > 1

, the correct value is

2

Step-by-Step Solution

Write the equations of lines

l_1

and

l_2

Line

l_1

has a slope of

m

and passes through

(0,0)

, so its equation is

y = mx

. Line

l_2

is perpendicular to

l_1

, so its slope is

-\frac{1}{m}

. Since its

y

-intercept is

(0,10)

, its equation is

y = -\frac{1}{m}x + 10

Setting up the equations of the lines allows us to find their point of intersection.

Find the

x

-coordinate of the intersection point

P

Equating the two expressions for

y

gives

mx = -\frac{1}{m}x + 10

. Multiplying both sides by

m

yields

m^2 x = -x + 10m

, which simplifies to

(m^2 + 1)x = 10m

, or

x = \frac{10m}{m^2 + 1}

The intersection point

P

must satisfy both equations simultaneously.

Solve for

m

using the distance from

P

to the

y

-axis.

The distance from

P(x,y)

to the

y

-axis is given by

|x|

. Since

m > 1

x

is positive, so the distance is

\frac{10m}{m^2 + 1} = 4

. This simplifies to

10m = 4m^2 + 4

, or

4m^2 - 10m + 4 = 0

. Dividing by

2

gives

2m^2 - 5m + 2 = 0

. Factoring the quadratic yields

(2m - 1)(m - 2) = 0

, giving solutions of

m = \frac{1}{2}

and

m = 2

. Since

m > 1

, the slope of

l_1

must be

2

Applying the given distance constraint and slope condition determines the unique value of

m

Key Concept

The relationship between the equations of perpendicular lines and their point of intersection in the coordinate plane.

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