Question:
Let
\[
A(1,2,1),\qquad
B\left(\frac95,\frac85,\frac95\right),\qquad
C(\alpha,\beta,\gamma)
\]
and
\[
D(-3,4,-3)
\]
be four collinear points. If \(B\) divides \(AC\) in the ratio \(m:n\), and \(D\) divides \(AC\) in the ratio \(m:-n\), then
\[
\alpha+\beta+\gamma=
\]
Let
\[
A(1,2,1),\qquad
B\left(\frac95,\frac85,\frac95\right),\qquad
C(\alpha,\beta,\gamma)
\]
and
\[
D(-3,4,-3)
\]
be four collinear points. If \(B\) divides \(AC\) in the ratio \(m:n\), and \(D\) divides \(AC\) in the ratio \(m:-n\), then
\[
\alpha+\beta+\gamma=
\]
Show Hint
When a point divides the same segment internally and another point divides it externally in the same ratio, introduce
\[
r=\frac{m}{n}
\]
to simplify all section-formula calculations.
Updated On: Jul 9, 2026
- \(7\)
- \(-2\)
- \(4\)
- \(\dfrac{26}{5}\) \bigskip
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
If a point divides a line segment internally in the ratio \(m:n\), then
\[
P=\left(
\frac{mx_2+nx_1}{m+n},
\frac{my_2+ny_1}{m+n},
\frac{mz_2+nz_1}{m+n}
\right).
\]
If it divides externally in the ratio \(m:n\), then
\[
P=\left(
\frac{mx_2-nx_1}{m-n},
\frac{my_2-ny_1}{m-n},
\frac{mz_2-nz_1}{m-n}
\right).
\]
Step 1: Let \(r=\dfrac{m}{n}\). Since \(B\) divides \(AC\) internally in the ratio \(m:n\), \[ B= \frac{mC+nA}{m+n} = \frac{rC+A}{r+1}. \] Using the \(x\)-coordinate, \[ \frac95 = \frac{r\alpha+1}{r+1}. \] \[ 9r+9=5r\alpha+5. \] \[ 5r\alpha=4r+4. \] \[ \alpha=\frac{4(r+1)}{5r}. \] \[ \cdots (1) \] Using the \(y\)-coordinate, \[ \frac85 = \frac{r\beta+2}{r+1}. \] \[ 8r+8=5r\beta+10. \] \[ 5r\beta=8r-2. \] \[ \beta=\frac{8r-2}{5r}. \] \[ \cdots (2) \] Using the \(z\)-coordinate, \[ \frac95 = \frac{r\gamma+1}{r+1}. \] Hence, \[ \gamma=\frac{4(r+1)}{5r}. \] \[ \cdots (3) \]
Step 2: Use the external division point \(D\). Since \(D\) divides \(AC\) in the ratio \(m:-n\), \[ D= \frac{mC-nA}{m-n} = \frac{rC-A}{r-1}. \] Using the \(x\)-coordinate, \[ -3 = \frac{r\alpha-1}{r-1}. \] Substituting (1), \[ -3 = \frac{\frac{4(r+1)}5-1}{r-1}. \] \[ -3 = \frac{4r-1}{5(r-1)}. \] \[ -15(r-1)=4r-1. \] \[ -15r+15=4r-1. \] \[ 19r=16. \] \[ r=\frac{16}{19}. \]
Step 3: Find \(\alpha,\beta,\gamma\). From (1), \[ \alpha = \frac{4\left(\frac{16}{19}+1\right)} {5\left(\frac{16}{19}\right)} = \frac74. \] Similarly, \[ \gamma=\frac74. \] From (2), \[ \beta = \frac{8\left(\frac{16}{19}\right)-2} {5\left(\frac{16}{19}\right)} = \frac72. \] Thus, \[ C = \left( \frac74,\frac72,\frac74 \right). \]
Step 4: Compute \(\alpha+\beta+\gamma\). \[ \alpha+\beta+\gamma = \frac74+\frac72+\frac74. \] \[ = \frac74+\frac74+\frac{14}{4}. \] \[ = \frac{28}{4}. \] \[ =7. \]
Step 5: Write the final answer. \[ \boxed{7} \]
Step 1: Let \(r=\dfrac{m}{n}\). Since \(B\) divides \(AC\) internally in the ratio \(m:n\), \[ B= \frac{mC+nA}{m+n} = \frac{rC+A}{r+1}. \] Using the \(x\)-coordinate, \[ \frac95 = \frac{r\alpha+1}{r+1}. \] \[ 9r+9=5r\alpha+5. \] \[ 5r\alpha=4r+4. \] \[ \alpha=\frac{4(r+1)}{5r}. \] \[ \cdots (1) \] Using the \(y\)-coordinate, \[ \frac85 = \frac{r\beta+2}{r+1}. \] \[ 8r+8=5r\beta+10. \] \[ 5r\beta=8r-2. \] \[ \beta=\frac{8r-2}{5r}. \] \[ \cdots (2) \] Using the \(z\)-coordinate, \[ \frac95 = \frac{r\gamma+1}{r+1}. \] Hence, \[ \gamma=\frac{4(r+1)}{5r}. \] \[ \cdots (3) \]
Step 2: Use the external division point \(D\). Since \(D\) divides \(AC\) in the ratio \(m:-n\), \[ D= \frac{mC-nA}{m-n} = \frac{rC-A}{r-1}. \] Using the \(x\)-coordinate, \[ -3 = \frac{r\alpha-1}{r-1}. \] Substituting (1), \[ -3 = \frac{\frac{4(r+1)}5-1}{r-1}. \] \[ -3 = \frac{4r-1}{5(r-1)}. \] \[ -15(r-1)=4r-1. \] \[ -15r+15=4r-1. \] \[ 19r=16. \] \[ r=\frac{16}{19}. \]
Step 3: Find \(\alpha,\beta,\gamma\). From (1), \[ \alpha = \frac{4\left(\frac{16}{19}+1\right)} {5\left(\frac{16}{19}\right)} = \frac74. \] Similarly, \[ \gamma=\frac74. \] From (2), \[ \beta = \frac{8\left(\frac{16}{19}\right)-2} {5\left(\frac{16}{19}\right)} = \frac72. \] Thus, \[ C = \left( \frac74,\frac72,\frac74 \right). \]
Step 4: Compute \(\alpha+\beta+\gamma\). \[ \alpha+\beta+\gamma = \frac74+\frac72+\frac74. \] \[ = \frac74+\frac74+\frac{14}{4}. \] \[ = \frac{28}{4}. \] \[ =7. \]
Step 5: Write the final answer. \[ \boxed{7} \]
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