JAMB 2014 Mathematics · Question 13

The gradient of a line joining (x,4) and (1,2) is 12\frac{1}{2}. Find the value of x

Options

  1. A.5
  2. B.3
  3. C.-3
  4. D.-5

Answer: A. 5

Explanation written with AI and checked against the answer key

Analysis

To find the value of xx, we need to use the formula for the gradient (slope) of a straight line passing through two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2).

Concept

The gradient (mm) of a line is defined as the change in yy divided by the change in xx:

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

In this problem, we are given:

  • Point 1: (x1,y1)=(x,4)(x_1, y_1) = (x, 4)
  • Point 2: (x2,y2)=(1,2)(x_2, y_2) = (1, 2)
  • Gradient (mm) = 12\frac{1}{2}
Solution
  1. Substitute the given values into the gradient formula: $12=2−41−x\frac{1}{2} = \frac{2 - 4}{1 - x}$
  1. Simplify the numerator: $12=−21−x\frac{1}{2} = \frac{-2}{1 - x}$
  1. Cross-multiply to solve for xx: $1(1−x)=2(−2)1(1 - x) = 2(-2)$ $1−x=−41 - x = -4$
  1. Isolate xx: $−x=−4−1-x = -4 - 1$ $−x=−5-x = -5$ $x=5x = 5$
Conclusion

By substituting the coordinates into the gradient formula and solving the resulting linear equation, we find that x=5x = 5.

Exam Tip: When dealing with gradient problems, always be careful with the signs of the coordinates. A common mistake is to subtract the coordinates in the wrong order (e.g., y1−y2y_1 - y_2 on top but x2−x1x_2 - x_1 on the bottom). Always maintain the same order for both: y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1} or y1−y2x1−x2\frac{y_1 - y_2}{x_1 - x_2}.

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