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Chapter 2: Problem 41

Find a number \(t\) such that the line in the \(x y\) -plane containing the points \((t, 4)\) and (2,-1) is perpendicular to the line \(y=6 x-7\).

Short Answer

Expert verified
The value of \(t\) that makes the line containing the points \((t, 4)\) and \((2, -1)\) perpendicular to the line \(y = 6x - 7\) is \(t = 28\).

Step by step solution

01

Find the slope of the line containing the points \((t, 4)\) and \((2, -1)\).

Using the formula for the slope of a line between two points \((x_1, y_1)\) and \((x_2, y_2)\), which is \(m = \frac{y_2 - y_1}{x_2 - x_1}\), we can find the slope of the line containing the points \((t, 4)\) and \((2, -1)\): \(m = \frac{-1 - 4}{2 - t}\).
02

Simplify the expression for the slope.

Simplify the expression for the slope: \(m = \frac{-5}{2 - t}\).
03

Find the slope of the given line \(y = 6x - 7\).

The given line is in slope-intercept form, which is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. Therefore, the slope of the given line is \(6\).
04

Determine the slope that makes the two lines perpendicular.

Two lines are perpendicular if their slopes are negative reciprocals of each other. Since the slope of the given line is \(6\), the slope of the line containing the points \((t, 4)\) and \((2, -1)\) should be the negative reciprocal of \(6\), which is \(\frac{-1}{6}\). Thus, we can set the value of the slope \(m\) found in Step 2 equal to this value: \(\frac{-5}{2 - t} = \frac{-1}{6}\).
05

Solve for \(t\).

To solve for \(t\), cross-multiply and solve the resulting equation: \(-5 \cdot 6 = -1 \cdot (2 - t)\) \(-30 = -2 + t\) Now, add \(30\) to both sides and add \(2\) to both sides: \(t = 28\).
06

State the solution.

The value of \(t\) that makes the line containing the points \((t, 4)\) and \((2, -1)\) perpendicular to the line \(y = 6x - 7\) is \(t = 28\).

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Most popular questions from this chapter

Write the indicated expression as a ratio of polynomials, assuming that $$ r(x)=\frac{3 x+4}{x^{2}+1}, \quad s(x)=\frac{x^{2}+2}{2 x-1}, \quad t(x)=\frac{5}{4 x^{3}+3} $$. $$ (r t)(x) $$

Suppose $$r(x)=\frac{x+1}{x^{2}+3} \quad \text { and } \quad s(x)=\frac{x+2}{x^{2}+5}$$ Find two distinct numbers \(x\) such that \(r(x)=\frac{1}{4}\).

Find a polynomial \(p\) of degree 3 such that \(-2,-1,\) and 4 are zeros of \(p\) and \(p(1)=2\).

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Find a number \(c\) such that -2 is a zero of the polynomial \(p\) defined by $$ p(x)=5-3 x+4 x^{2}+c x^{3} $$.
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