A straight line through the point \[A(3,4)\] is such that its intercept between the axes is bisected at A. it’s equation is-
A. \[3x - 4y + 7 = 0\]
B. \[4x+ 3y = 24\]
C. \[3x - 4y = 25\]
D. \[x - y = 7\]
Answer
633.9k+ views
Hint: In this question, the point \[A(3,4)\] which passes through the straight line. To find the equation of line, we use the slope-intercept formula. First we determine the end point of the straight line and evaluate by using the formula.
Complete step-by-step answer:
Here \[A(3,4)\] is a point which passes through the straight line. Consider \[BC\] is a straight-line having points \[(a,0)\] and \[(0,b)\] . Let \[A\] be the midpoint of the line \[BC\] .
By using the midpoint formula, we determine the end points as
\[
\dfrac{{a + 0}}{2} = 3{\text{ and }}\dfrac{{0 + b}}{2} = 4 \\
\Rightarrow \dfrac{a}{2} = 3{\text{ and }}\dfrac{b}{2} = 4 \;
\]
By simplifying, we have \[a = 6\] and \[b = 8\] .
Hence, we determined the endpoints. Now we have to find the equation of line.
To find the equation of line we consider the slope-intercept formula which is given by \[y = mx + b\] . \[y\] is \[y\] coordinate, \[m\] is slope, \[x\] is \[x\] coordinate and \[b\] is \[y\] intercept.
First we determine the slope \[m\] is given by \[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\] where \[({x_1},{y_1})\] and \[({x_2},{y_2})\] are the endpoints of the straight line. Here \[({x_1},{y_1})\] and \[({x_2},{y_2})\] are \[(6,0)\] and \[(0,8)\]
\[
m = \dfrac{{8 - 0}}{{0 - 6}} \\
= \dfrac{8}{{ - 6}} = - \dfrac{8}{6} \;
\]
Therefore the equation of line by intercept formula is \[y = mx + b\]
\[y = \left( { - \dfrac{8}{6}} \right)x + 8\]
By taking L.C.M.
\[
y = \dfrac{{ - 8x + 48}}{6} \\
\Rightarrow 6y = - 8x + 48 \;
\]
Dividing the above equation by 2 we have
\[
3y = - 4x + 24 \\
\Rightarrow 4x + 3y = 24 \;
\]
∴ the equation of line is given by \[4x + 3y = 24\]
So, the correct answer is “Option B”.
Note: To find the equation of a line we need the end points. In this question first, we have to determine the endpoints of the straight line. After determining the endpoints, we can determine the equation of the line by using the slope intercept formula. Here we considered the point \[A(3,4)\] as midpoint and we determined the end points and evaluated them using the slope intercept formula.
Complete step-by-step answer:
Here \[A(3,4)\] is a point which passes through the straight line. Consider \[BC\] is a straight-line having points \[(a,0)\] and \[(0,b)\] . Let \[A\] be the midpoint of the line \[BC\] .
By using the midpoint formula, we determine the end points as
\[
\dfrac{{a + 0}}{2} = 3{\text{ and }}\dfrac{{0 + b}}{2} = 4 \\
\Rightarrow \dfrac{a}{2} = 3{\text{ and }}\dfrac{b}{2} = 4 \;
\]
By simplifying, we have \[a = 6\] and \[b = 8\] .
Hence, we determined the endpoints. Now we have to find the equation of line.
To find the equation of line we consider the slope-intercept formula which is given by \[y = mx + b\] . \[y\] is \[y\] coordinate, \[m\] is slope, \[x\] is \[x\] coordinate and \[b\] is \[y\] intercept.
First we determine the slope \[m\] is given by \[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\] where \[({x_1},{y_1})\] and \[({x_2},{y_2})\] are the endpoints of the straight line. Here \[({x_1},{y_1})\] and \[({x_2},{y_2})\] are \[(6,0)\] and \[(0,8)\]
\[
m = \dfrac{{8 - 0}}{{0 - 6}} \\
= \dfrac{8}{{ - 6}} = - \dfrac{8}{6} \;
\]
Therefore the equation of line by intercept formula is \[y = mx + b\]
\[y = \left( { - \dfrac{8}{6}} \right)x + 8\]
By taking L.C.M.
\[
y = \dfrac{{ - 8x + 48}}{6} \\
\Rightarrow 6y = - 8x + 48 \;
\]
Dividing the above equation by 2 we have
\[
3y = - 4x + 24 \\
\Rightarrow 4x + 3y = 24 \;
\]
∴ the equation of line is given by \[4x + 3y = 24\]
So, the correct answer is “Option B”.
Note: To find the equation of a line we need the end points. In this question first, we have to determine the endpoints of the straight line. After determining the endpoints, we can determine the equation of the line by using the slope intercept formula. Here we considered the point \[A(3,4)\] as midpoint and we determined the end points and evaluated them using the slope intercept formula.
Recently Updated Pages
Difference Between Prokaryotic Cells and Eukaryotic Cells

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

