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NCERT Exemplar · Q17

Q.Let A(2,2,−3)A(2,2,-3), B(5,6,9)B(5,6,9) and C(2,7,9)C(2,7,9) be the vertices of a triangle. The internal bisector of the angle A meets BC at the point D. Find the coordinates of D.

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The internal angle bisector of angle A divides the opposite side BC in the ratio of the adjacent sides AB and AC. By calculating the lengths ABAB and ACAC, we find they are equal, meaning D is the midpoint of BC. The coordinates of D are (72,132,9)\boxed{\left(\frac{7}{2}, \frac{13}{2}, 9\right)}.

The problem asks for the coordinates of point D, where the internal bisector of angle A of triangle ABC meets side BC. This is a classic application of the Angle Bisector Theorem, which holds true in three-dimensional coordinate geometry just as it does in two dimensions.

The core idea is that an angle bisector divides the opposite side in a ratio equal to the ratio of the other two sides of the triangle. Specifically, if AD is the internal bisector of angle A, then point D divides the side BC in the ratio AB:ACAB:AC. Once we find this ratio, we can use the section formula to determine the coordinates of D.

  1. Understand the given information and the goal.

    We are given the coordinates of the three vertices of a triangle: A(2,2,−3)A(2,2,-3), B(5,6,9)B(5,6,9), and C(2,7,9)C(2,7,9). We need to find the coordinates of point D, which lies on BC and is the point where the internal bisector of angle A intersects BC.

  2. Recall the Angle Bisector Theorem.

    The Angle Bisector Theorem states that if a line internally bisects an angle of a triangle, then it divides the opposite side into two segments that are proportional to the other two sides of the triangle.

    In △ABC\triangle ABC, if AD is the internal bisector of ∠A\angle A, then:

BDDC=ABAC\frac{BD}{DC} = \frac{AB}{AC}

This means that point D divides the line segment BC internally in the ratio $AB:AC$. Let this ratio be $m:n$, where $m = AB$ and $n = AC$.

3. Calculate the lengths of sides AB and AC.

We use the distance formula in 3D for two points (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2):

> [!FORMULA]

> The distance between two points (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2) is given by:

> d=(x2−x1)2+(y2−y1)2+(z2−z1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

*   **Length of AB:** Using $A(2,2,-3)$ and $B(5,6,9)$:

AB=(5−2)2+(6−2)2+(9−(−3))2AB = \sqrt{(5-2)^2 + (6-2)^2 + (9-(-3))^2}

AB=(3)2+(4)2+(12)2AB = \sqrt{(3)^2 + (4)^2 + (12)^2}

AB=9+16+144AB = \sqrt{9 + 16 + 144}

AB=169AB = \sqrt{169}

AB=13AB = 13

*   **Length of AC:** Using $A(2,2,-3)$ and $C(2,7,9)$:

AC=(2−2)2+(7−2)2+(9−(−3))2AC = \sqrt{(2-2)^2 + (7-2)^2 + (9-(-3))^2}

AC=(0)2+(5)2+(12)2AC = \sqrt{(0)^2 + (5)^2 + (12)^2}

AC=0+25+144AC = \sqrt{0 + 25 + 144}

AC=169AC = \sqrt{169}

AC=13AC = 13

  1. Determine the ratio in which D divides BC. From the Angle Bisector Theorem, DD divides BCBC in the ratio AB:ACAB:AC. We found AB=13AB = 13 and AC=13AC = 13. So, the ratio m:n=AB:AC=13:13=1:1m:n = AB:AC = 13:13 = 1:1. …

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