Two lines in a coordinate plane, T_1 and T_2, are perpendicular to each other. Line T_1 ha… — ACT

Two lines in a coordinate plane, $T_1$ and $T_2$ , are perpendicular to each other. Line $T_1$ has the equation $3x - 5y = 15$ , and line $T_2$ has the equation $ax + 9y = 20$ , where $a$ is a constant. What is the value of $a$ ?

Answer: 15

Answer

The value of the constant

a

is 15.

The correct answer is 15. The slope of the line

3x - 5y = 15

\frac{3}{5}

. Since the two lines are perpendicular, the slope of the second line must be the negative reciprocal of

\frac{3}{5}

, which is

-\frac{5}{3}

. The slope of the line

ax + 9y = 20

-\frac{a}{9}

. Equating the two slopes yields

-\frac{a}{9} = -\frac{5}{3}

, which simplifies to

a = 15

Step-by-Step Solution

Find the slope of line

T_1

The slope of line

T_1

\frac{3}{5}

Writing the equation

3x - 5y = 15

in slope-intercept form

y = \frac{3}{5}x - 3

isolates the slope coefficient.

Find the perpendicular slope

The perpendicular slope is

-\frac{5}{3}

Perpendicular lines have slopes that are negative reciprocals of each other, so we invert the fraction and change the sign.

Determine the slope of line

T_2

in terms of

a

The slope of line

T_2

-\frac{a}{9}

Rewriting the equation

ax + 9y = 20

in slope-intercept form

y = -\frac{a}{9}x + \frac{20}{9}

isolates the slope coefficient.

Solve for

a

a = 15

Setting the slope of

T_2

equal to the perpendicular slope yields the equation

-\frac{a}{9} = -\frac{5}{3}

, which simplifies to

a = 15

Key Concept

The slopes of perpendicular lines are negative reciprocals of each other.

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