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Question: If a point C with y coordinate 2, lies on the line joining the points A (-1, -4, 5) and B (4, 6, -5)...

If a point C with y coordinate 2, lies on the line joining the points A (-1, -4, 5) and B (4, 6, -5), then find the coordinate of C.
A) (2, 2, -1)
B) (2, 1, 2)
C) (2, 0, 0)
D) (0, 2, 4)

Explanation

Solution

Here first we will find the equation of line AB and then satisfy the y coordinate of C and find the x and z coordinates then.
The equation of a line passing through two points (x1,y1,z1)\left( {{x_1},{y_1},{z_1}} \right) and (x2,y2,z2)\left( {{x_2},{y_2},{z_2}} \right) is given by:-
x−x2x1−x2=y−y2y1−y2=z−z2z1−z2\dfrac{{x - {x_2}}}{{{x_1} - {x_2}}} = \dfrac{{y - {y_2}}}{{{y_1} - {y_2}}} = \dfrac{{z - {z_2}}}{{{z_1} - {z_2}}}

Complete step-by-step answer:
The given endpoints of the line are:-
A (-1, 4, 5) and B (4, 6, -5)
Now we know that:-
The equation of a line passing through two points (x1,y1,z1)\left( {{x_1},{y_1},{z_1}} \right) and (x2,y2,z2)\left( {{x_2},{y_2},{z_2}} \right) is given by:-
x−x2x1−x2=y−y2y1−y2=z−z2z1−z2\dfrac{{x - {x_2}}}{{{x_1} - {x_2}}} = \dfrac{{y - {y_2}}}{{{y_1} - {y_2}}} = \dfrac{{z - {z_2}}}{{{z_1} - {z_2}}}
So for the above points
x1=−1,y1=−4,z1=5{x_1} = - 1,{y_1} = - 4,{z_1} = 5
And, x2=4,y2=6,z2=−5{x_2} = 4,{y_2} = 6,{z_2} = - 5
Therefore, applying the formula we get:-
The equation of line AB is given by:-
x−4−1−4=y−6−4−6=z−(−5)5−(−5)\dfrac{{x - 4}}{{ - 1 - 4}} = \dfrac{{y - 6}}{{ - 4 - 6}} = \dfrac{{z - ( - 5)}}{{5 - ( - 5)}}
Solving it further we get:-
x−4−5=y−6−10=z+510\dfrac{{x - 4}}{{ - 5}} = \dfrac{{y - 6}}{{ - 10}} = \dfrac{{z + 5}}{{10}}
Putting this equation equal to constant k we get:-
x−4−5=y−6−10=z+510=k\dfrac{{x - 4}}{{ - 5}} = \dfrac{{y - 6}}{{ - 10}} = \dfrac{{z + 5}}{{10}} = k
Now equating first and fourth term we get:-
x−4−5=k\dfrac{{x - 4}}{{ - 5}} = k
Solving for x we get:-
x−4=−5kx - 4 = - 5k
⇒x=−5k+4\Rightarrow x = - 5k + 4…………………………… (1)
Now equating third and fourth term we get:-
z+510=k\dfrac{{z + 5}}{{10}} = k
Solving for z we get:-
z+5=10kz + 5 = 10k
⇒z=10k−5\Rightarrow z = 10k - 5……………………………….. (2)
Now equating second and fourth term we get:-
y−6−10=k\dfrac{{y - 6}}{{ - 10}} = k
Solving for y we get:-
y−6=−10ky - 6 = - 10k
⇒y=−10k+6\Rightarrow y = - 10k + 6………………………….. (3)
Now we since point C lies on the line AB
Therefore, the coordinates of C should satisfy the equation of AB
Now since we know that y coordinate of C is 2
Hence substituting this value in equation 3 we get:-
2=−10k+62 = - 10k + 6
Solving for k we get:-
−10k=2−6- 10k = 2 - 6
⇒k=−4−10\Rightarrow k = \dfrac{{ - 4}}{{ - 10}}
⇒k=25\Rightarrow k = \dfrac{2}{5}
Putting this value in equation1 we get:-
x=−5(25)+4x = - 5\left( {\dfrac{2}{5}} \right) + 4
Solving for x we get:-

⇒x=−2+4 ⇒x=2  \Rightarrow x = - 2 + 4 \\\ \Rightarrow x = 2 \\\

Now putting the value of k in equation 2 we get:-
z=10(25)−5z = 10\left( {\dfrac{2}{5}} \right) - 5
Solving for z we get:-

z=4−5 ⇒z=−1  z = 4 - 5 \\\ \Rightarrow z = - 1 \\\

Therefore the coordinates of C are (2, 2, -1)

Hence option A is correct.

Note: Students should take a note that when a point lies on a curve or a line then its coordinates should satisfy the equation of the curve or the line.