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Difficulty: HardLinear Equations in Two Variables

In the xyxy-plane, a line with a positive slope mm passes through the point (4,−3)(4, -3) and intersects the xx-axis at (p,0)(p, 0) and the yy-axis at (0,q)(0, q), where pp and qq are non-zero constants. If p+q=5p + q = 5, what is the value of mm?

Answer: 0.5

Answer

The value of mm is 0.50.5 (or 1/21/2).
To find the slope mm, write the equation of the line in point-slope form: y−(−3)=m(x−4)y - (-3) = m(x - 4), which simplifies to y=mx−4m−3y = mx - 4m - 3. The yy-intercept is found by setting x=0x = 0, giving q=−4m−3q = -4m - 3. The xx-intercept is found by setting y=0y = 0, giving p=4+3mp = 4 + \frac{3}{m}. Substituting these expressions into p+q=5p + q = 5 gives 4+3m−4m−3=54 + \frac{3}{m} - 4m - 3 = 5, which simplifies to −4m−4+3m=0-4m - 4 + \frac{3}{m} = 0. Multiplying this equation by −m-m yields the quadratic equation 4m2+4m−3=04m^2 + 4m - 3 = 0. Factoring the quadratic gives (2m−1)(2m+3)=0(2m - 1)(2m + 3) = 0. Since the line is defined to have a positive slope, the value of mm must be positive, which is 0.50.5 (or 1/21/2).

Step-by-Step Solution

1
Write the general equation of a line passing through (4,−3)(4, -3) with slope mm.
y=mx−4m−3y = mx - 4m - 3
Using the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with (x1,y1)=(4,−3)(x_1, y_1) = (4, -3) establishes the relationship between xx, yy, and mm.
2
Determine the value of qq by finding the yy-intercept of the line.
q=−4m−3q = -4m - 3
The yy-intercept (0,q)(0, q) occurs when x=0x = 0.
3
Determine the value of pp by finding the xx-intercept of the line.
p=4+3mp = 4 + \frac{3}{m}
The xx-intercept (p,0)(p, 0) occurs when y=0y = 0.
4
Substitute the expressions for pp and qq into the given equation p+q=5p + q = 5.
4+3m−4m−3=54 + \frac{3}{m} - 4m - 3 = 5
This sets up a single equation with the variable mm using the given relationship.
5
Simplify the equation and rewrite it in standard quadratic form.
4m2+4m−3=04m^2 + 4m - 3 = 0
Multiplying the simplified expression −4m−4+3m=0-4m - 4 + \frac{3}{m} = 0 by −m-m eliminates the fraction and aligns it to standard quadratic form.
6
Factor the quadratic equation to find the possible values of mm.
m=0.5m = 0.5 or m=−1.5m = -1.5
Factoring (2m−1)(2m+3)=0(2m - 1)(2m + 3) = 0 gives the roots of the equation.
7
Filter the results based on the positive slope constraint.
m=0.5m = 0.5
The problem states that the line has a positive slope, so we must reject the negative root.

Key Concept

Using linear intercepts to solve system constraints

Alternative Method

Instead of using point-slope form, use the intercept form of a linear equation: xp+yq=1\frac{x}{p} + \frac{y}{q} = 1. Since the line passes through the point (4,−3)(4, -3), substitute x=4x = 4 and y=−3y = -3 to get 4p−3q=1\frac{4}{p} - \frac{3}{q} = 1. Given p+q=5p + q = 5, substitute p=5−qp = 5 - q into the equation to get 45−q−3q=1\frac{4}{5 - q} - \frac{3}{q} = 1. Multiply both sides by the common denominator q(5−q)q(5 - q) to obtain 4q−3(5−q)=q(5−q)  ⟹  7q−15=5q−q2  ⟹  q2+2q−15=04q - 3(5 - q) = q(5 - q) \implies 7q - 15 = 5q - q^2 \implies q^2 + 2q - 15 = 0. Factoring gives (q+5)(q−3)=0(q + 5)(q - 3) = 0. This yields q=−5q = -5 or q=3q = 3. If q=3q = 3, then p=2p = 2, and the slope is m=−qp=−32m = -\frac{q}{p} = -\frac{3}{2} (negative). If q=−5q = -5, then p=10p = 10, and the slope is m=−qp=−−510=0.5m = -\frac{q}{p} = -\frac{-5}{10} = 0.5 (positive). Since the slope must be positive, the slope of the line is 0.50.5.
Estimated Time:2m 30s
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Topics

Linear Equations in Two Variables

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