Collinearity of points MCQ Quiz in मल्याळम - Objective Question with Answer for Collinearity of points - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT:

If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then \(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}}\\ {{x_2}}&{{y_2}}&{{z_2}}\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right| = 0\)

CONCEPT:

If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then \(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}}\\ {{x_2}}&{{y_2}}&{{z_2}}\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right| = 0\)

Given:

Points: (-5, 1), (1, k), and (4, -2)

Formula used:

For three points to be collinear, the area of the triangle formed by them must be zero.

Area of a triangle using coordinates:

Area = (1/2) × [x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

Calculation:

Since the points are collinear:

(1/2) × [-5(k + 2) + 1(-2 - 1) + 4(1 - k)] = 0

-5(k + 2) + 1(-3) + 4(1 - k) = 0

-5k - 10 - 3 + 4 - 4k = 0

-9k - 9 = 0

-9k = 9

k = -1

∴ The value of k is -1.

\(\vec{PR}=(5-4)\hat{i}+(8-5)\hat{j}+(0-x)\hat{k}\)

\(\implies \vec{PR}=\hat{i}+3\hat{j}-x\hat{k}\)

Again, \(\vec{QR}=(5-3)\hat{i}+(8-y)\hat{j}+(0-4)\hat{k}\)

\(\implies \vec{QR}=2\hat{i}+(8-y)\hat{j}-4\hat{k}\)

Since, P, Q and R are co-linear points, hence

\(\dfrac{1}{2}=\dfrac{3}{8-y}=\dfrac{-x}{-4}\)

On Solving, we get:-

\(x=2,y=2\)

\(\implies x+y=4\)

Hence, answer is option-(D).