QUESTION IMAGE
Question
section 2 each question carries 3 mark (3*3=9)
- p (3,1) and q (-1, a) are 5 units apart. find a
Step1: Recall distance formula
The distance \(d\) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here, \(P(3,1)\), \(Q(-1,a)\), and \(d = 5\).
Step2: Substitute values into formula
Substitute \(x_1 = 3\), \(y_1 = 1\), \(x_2=-1\), \(y_2 = a\), and \(d = 5\) into the formula:
\(5=\sqrt{(-1 - 3)^2+(a - 1)^2}\)
Simplify inside the square root: \(-1-3=-4\), so \((-4)^2 = 16\). The equation becomes \(5=\sqrt{16+(a - 1)^2}\).
Step3: Square both sides
Square both sides to eliminate the square root: \(5^2=(\sqrt{16+(a - 1)^2})^2\), which gives \(25 = 16+(a - 1)^2\).
Step4: Solve for \((a - 1)^2\)
Subtract 16 from both sides: \(25-16=(a - 1)^2\), so \(9=(a - 1)^2\).
Step5: Solve for \(a\)
Take square roots of both sides: \(a - 1=\pm3\).
Case 1: \(a - 1 = 3\), then \(a=3 + 1=4\).
Case 2: \(a - 1=-3\), then \(a=-3 + 1=-2\).
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\(a = 4\) or \(a=-2\)