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Chapter 2: Problem 43

Find a number \(t\) such that the line containing the points \((4, t)\) and (-1,6) is perpendicular to the line that contains the points (3,5) and (1,-2) .

Short Answer

Expert verified
The value of t for which the line containing the points \((4, t)\) and \((-1, 6)\) is perpendicular to the line containing the points \((3, 5)\) and \((1, -2)\) is \(t = \cfrac{32}{7}\).

Step by step solution

01

Find the slope of the line containing the points (3,5) and (1,-2)

To find the slope, we can use the slope formula: \( slope (m) = \cfrac{ y_2 - y_1}{x_2 - x_1}\) For the points (3,5) as (x1, y1) and (1,-2) as (x2, y2), we can substitute into the formula and calculate: \(m = \cfrac{-2 - 5}{1 - 3} = \cfrac{-7}{-2} = \cfrac{7}{2}\) The slope of the line containing the points (3,5) and (1,-2) is \(m = \cfrac{7}{2}\).
02

Determine the slope of the perpendicular line

The slope of a line perpendicular to another line with slope 'm' is the negative reciprocal of that slope. For the given slope \(\cfrac{7}{2}\), the negative reciprocal can be found as follows: \( m_{perpendicular} = -\cfrac{1}{m} = -\cfrac{1}{\cfrac{7}{2}} = -\cfrac{2}{7} \) The slope of the line perpendicular to the given line is \(m_{perpendicular} = -\cfrac{2}{7}\).
03

Set up the general equation for the line containing points (4, t) and (-1,6)

Using the slope-point form of the line, we can write the equation for the line containing points (4, t) and (-1,6) in terms of the slope, m_{perpendicular}, and the coordinates of the points: \( y - t = m_{perpendicular}(x - 4) \) We know the value of \(m_{perpendicular}\) is \(-\cfrac{2}{7}\), so substituting it, we get: \( y - t = -\cfrac{2}{7}(x - 4) \)
04

Plug in the coordinates of (-1,6) into the equation and solve for t

Now, we can substitute the coordinates of the other point (-1, 6) into the equation and solve for t: \(6 - t = -\cfrac{2}{7}(-1 - 4)\) Next, we simplify the equation: \(6 - t = -\cfrac{2}{7}(-5)\) and, \(6 - t = \cfrac{10}{7}\) Then, we solve for t: \(t = 6 - \cfrac{10}{7}\) To find t, we express 6 as a fraction with denominator of 7: \(t = \cfrac{42}{7} - \cfrac{10}{7}\) Finally, we subtract the fractions: \(t = \cfrac{32}{7}\)
05

State the result

The value of t is \(\cfrac{32}{7}\), so the y-coordinate of the point (4, t) that makes the line containing points (4, t) and (-1, 6) perpendicular to the line containing points (3,5) and (1,-2) is \(t = \cfrac{32}{7}\).

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Most popular questions from this chapter

Suppose \(p(x)=a_{0}+a_{1} x+\cdots+a_{n} x^{n},\) where \(a_{0}, a_{1}, \ldots, a_{n}\) are integers. Suppose \(M\) and \(N\) are nonzero integers with no common factors and \(\frac{M}{N}\) is a zero of \(p\). Show that \(a_{0} / M\) and \(a_{n} / N\) are integers. [Thus to find rational zeros of a polynomial with integer coefficients, we need only look at fractions whose numerator is a divisor of the constant term and whose denominator is a divisor of the coefficient of highest degree. This result is called the Rational Zeros Theorem or the Rational Roots Theorem.]

Suppose \(M\) and \(N\) are odd integers. Explain why $$ x^{2}+M x+N $$ has no rational zeros.

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