
The equation of line is \[ - 3y + 4x = 9\]. How do you write the equation of a line that is parallel to the line and passes through the point \[\left( { - 12,6} \right)\]?
Answer
523.8k+ views
Hint: Two lines are either intersecting at one point or never intersect at any point (parallel lines). If two lines are parallel then their slopes are equal and intercepts are different in the slope intercept form of a line.
Complete step by step solution:
The given equation of a line is \[ - 3y + 4x = 9\].
Write the given equation in a slope intercept form \[y = mx + c\] where \[m\] represent slope of a line and \[c\] represent the y-intercept.
Subtract \[4x\] from both sides of the given equation as shown below.
\[ - 3y + 4x - 4x = 9 - 4x\]
\[ \Rightarrow - 3y = 9 - 4x\]
Now, divide by \[ - 3\] to both sides of the above equation as follows:
\[ \Rightarrow \dfrac{{ - 3y}}{{ - 3}} = \dfrac{{9 - 4x}}{{ - 3}}\]
Simplify the equation as shown below.
\[ \Rightarrow y = - \dfrac{9}{3} + \dfrac{4}{3}x\]
\[ \Rightarrow y = - 3 + \dfrac{4}{3}x\]
\[ \Rightarrow y = \dfrac{4}{3}x - 3\]
Comparing the above equation with slope intercept form, it is observed that slope of line is \[\dfrac{4}{3}\] and y-intercept is \[ - 3\].
Now, to form an equation of a line that is parallel to given equation of a line and passes through the point \[\left( { - 12,6} \right)\] we have to use point slope form that is the equation of a line with slope \[m\] and passes through the point \[\left( {{x_1},{y_1}} \right)\] is \[y - {y_1} = m\left( {x - {x_1}} \right)\].
As we know parallel lines have the same slope, therefore the slope of a required parallel line equation is \[\dfrac{4}{3}\] but it passes through the point \[\left( { - 12,6} \right)\].
Substitute \[m\] as \[\dfrac{4}{3}\] and \[\left( {{x_1},{y_1}} \right)\] as \[\left( { - 12,6} \right)\] in the equation \[y - {y_1} = m\left( {x - {x_1}} \right)\] as shown below.
\[y - \left( 6 \right) = \left( {\dfrac{4}{3}} \right)\left( {x - \left( { - 12} \right)} \right)\]
Simplify the equation as shown below.
\[ \Rightarrow y - 6 = \left( {\dfrac{4}{3}} \right)\left( {x + 12} \right)\]
Multiply both sides by number \[3\] and simplify further.
\[ \Rightarrow 3\left( {y - 6} \right) = 4\left( {x + 12} \right)\]
\[ \Rightarrow 3y - 18 = 4x + 48\]
Rearrange the equation to write it in standard form as shown below.
\[ \Rightarrow 4x - 3y + 18 + 48 = 0\]
\[ \Rightarrow 4x - 3y + 66 = 0\]
Therefore, the required equation of a line that is parallel to the line of an equation\[ - 3y + 4x = 9\] and passes through the point \[\left( { - 12,6} \right)\] is \[4x - 3y + 66 = 0\].
Note: The ratio of coefficient of \[x\] is equal to the ratio of coefficient of \[y\] and not equal to the ratio of constant numbers for two parallel lines. Similarly the ratio of coefficient of \[x\] is not equal to the ratio of coefficient of \[y\] for two intersecting lines.
Complete step by step solution:
The given equation of a line is \[ - 3y + 4x = 9\].
Write the given equation in a slope intercept form \[y = mx + c\] where \[m\] represent slope of a line and \[c\] represent the y-intercept.
Subtract \[4x\] from both sides of the given equation as shown below.
\[ - 3y + 4x - 4x = 9 - 4x\]
\[ \Rightarrow - 3y = 9 - 4x\]
Now, divide by \[ - 3\] to both sides of the above equation as follows:
\[ \Rightarrow \dfrac{{ - 3y}}{{ - 3}} = \dfrac{{9 - 4x}}{{ - 3}}\]
Simplify the equation as shown below.
\[ \Rightarrow y = - \dfrac{9}{3} + \dfrac{4}{3}x\]
\[ \Rightarrow y = - 3 + \dfrac{4}{3}x\]
\[ \Rightarrow y = \dfrac{4}{3}x - 3\]
Comparing the above equation with slope intercept form, it is observed that slope of line is \[\dfrac{4}{3}\] and y-intercept is \[ - 3\].
Now, to form an equation of a line that is parallel to given equation of a line and passes through the point \[\left( { - 12,6} \right)\] we have to use point slope form that is the equation of a line with slope \[m\] and passes through the point \[\left( {{x_1},{y_1}} \right)\] is \[y - {y_1} = m\left( {x - {x_1}} \right)\].
As we know parallel lines have the same slope, therefore the slope of a required parallel line equation is \[\dfrac{4}{3}\] but it passes through the point \[\left( { - 12,6} \right)\].
Substitute \[m\] as \[\dfrac{4}{3}\] and \[\left( {{x_1},{y_1}} \right)\] as \[\left( { - 12,6} \right)\] in the equation \[y - {y_1} = m\left( {x - {x_1}} \right)\] as shown below.
\[y - \left( 6 \right) = \left( {\dfrac{4}{3}} \right)\left( {x - \left( { - 12} \right)} \right)\]
Simplify the equation as shown below.
\[ \Rightarrow y - 6 = \left( {\dfrac{4}{3}} \right)\left( {x + 12} \right)\]
Multiply both sides by number \[3\] and simplify further.
\[ \Rightarrow 3\left( {y - 6} \right) = 4\left( {x + 12} \right)\]
\[ \Rightarrow 3y - 18 = 4x + 48\]
Rearrange the equation to write it in standard form as shown below.
\[ \Rightarrow 4x - 3y + 18 + 48 = 0\]
\[ \Rightarrow 4x - 3y + 66 = 0\]
Therefore, the required equation of a line that is parallel to the line of an equation\[ - 3y + 4x = 9\] and passes through the point \[\left( { - 12,6} \right)\] is \[4x - 3y + 66 = 0\].
Note: The ratio of coefficient of \[x\] is equal to the ratio of coefficient of \[y\] and not equal to the ratio of constant numbers for two parallel lines. Similarly the ratio of coefficient of \[x\] is not equal to the ratio of coefficient of \[y\] for two intersecting lines.
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success

Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

