
How do you write an equation in slope intercept form given two points?
Answer
469.5k+ views
Hint: We first find the slope-intercept form for any point $P\left( x,y \right)$ and also find the equation of the line. In this case we actually take the help of the origin where $O\left( 0,0 \right)$. Then we take two arbitrary points and place them on the equation of the line and find the value of the slope and its equation.
Complete step-by-step solution:
We need to find the equation of the line which is inclined at a given angle with the positive direction of the axis of x and cuts off a given intercept from the axis of y.
Suppose the line $\overleftrightarrow{AB}$ intersects the X-axis at D and the y-axis at C. if the line makes an angle $\alpha $ with the positive direction of X-axis and $\overleftrightarrow{OC}=c$. We have to find the equation of the line $\overleftrightarrow{AB}$.
Let $P\left( x,y \right)$ be any point on the line $\overleftrightarrow{AB}$. We draw perpendicular $\overleftrightarrow{PM}$ on $\overleftrightarrow{OX}$ and $\overleftrightarrow{CE}$ perpendicular on $\overleftrightarrow{PM}$. Since, $CE||DM$, we have $\angle PCE=\angle CDM=\alpha $.
We also have $\overleftrightarrow{PM}=y;\overleftrightarrow{OC}=c;\overleftrightarrow{CE}=\overleftrightarrow{OM}=x$.
Then we have $\overleftrightarrow{PE}=\overleftrightarrow{PM}-\overleftrightarrow{EM}=\overleftrightarrow{PM}-\overleftrightarrow{OC}=y-c$.
Therefore, from the right-angled $\Delta PCE$, we get $\tan \alpha =\dfrac{\overleftrightarrow{PE}}{\overleftrightarrow{CE}}=\dfrac{y-c}{x}$.
We assume that the slope is m where $m=\tan \alpha $ which gives $m=\tan \alpha =\dfrac{y-c}{x}$.
The equation becomes $y=mx+c$.
Now if two points are given of the line where $P\left( x,y \right)$ and $Q\left( {{x}_{1}},{{y}_{1}} \right)$, the only change happens in the slope. For the above problem we took the second point as the origin.
Now both $P\left( x,y \right)$ and $Q\left( {{x}_{1}},{{y}_{1}} \right)$ goes through $y=mx+c$.
This gives $y=mx+c$ and ${{y}_{1}}=m{{x}_{1}}+c$. Subtracting we get $y-{{y}_{1}}=mx-m{{x}_{1}}=m\left( x-{{x}_{1}} \right)$.
The value of m becomes $m=\dfrac{\left( y-{{y}_{1}} \right)}{\left( x-{{x}_{1}} \right)}$. The equation becomes $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$.
Note: We could have used the same process where we replace the origin value $O\left( 0,0 \right)$ with the point $Q\left( {{x}_{1}},{{y}_{1}} \right)$. The shifting of the origin will be sufficed to do that.
Complete step-by-step solution:
We need to find the equation of the line which is inclined at a given angle with the positive direction of the axis of x and cuts off a given intercept from the axis of y.
Suppose the line $\overleftrightarrow{AB}$ intersects the X-axis at D and the y-axis at C. if the line makes an angle $\alpha $ with the positive direction of X-axis and $\overleftrightarrow{OC}=c$. We have to find the equation of the line $\overleftrightarrow{AB}$.

Let $P\left( x,y \right)$ be any point on the line $\overleftrightarrow{AB}$. We draw perpendicular $\overleftrightarrow{PM}$ on $\overleftrightarrow{OX}$ and $\overleftrightarrow{CE}$ perpendicular on $\overleftrightarrow{PM}$. Since, $CE||DM$, we have $\angle PCE=\angle CDM=\alpha $.
We also have $\overleftrightarrow{PM}=y;\overleftrightarrow{OC}=c;\overleftrightarrow{CE}=\overleftrightarrow{OM}=x$.
Then we have $\overleftrightarrow{PE}=\overleftrightarrow{PM}-\overleftrightarrow{EM}=\overleftrightarrow{PM}-\overleftrightarrow{OC}=y-c$.
Therefore, from the right-angled $\Delta PCE$, we get $\tan \alpha =\dfrac{\overleftrightarrow{PE}}{\overleftrightarrow{CE}}=\dfrac{y-c}{x}$.
We assume that the slope is m where $m=\tan \alpha $ which gives $m=\tan \alpha =\dfrac{y-c}{x}$.
The equation becomes $y=mx+c$.
Now if two points are given of the line where $P\left( x,y \right)$ and $Q\left( {{x}_{1}},{{y}_{1}} \right)$, the only change happens in the slope. For the above problem we took the second point as the origin.
Now both $P\left( x,y \right)$ and $Q\left( {{x}_{1}},{{y}_{1}} \right)$ goes through $y=mx+c$.
This gives $y=mx+c$ and ${{y}_{1}}=m{{x}_{1}}+c$. Subtracting we get $y-{{y}_{1}}=mx-m{{x}_{1}}=m\left( x-{{x}_{1}} \right)$.
The value of m becomes $m=\dfrac{\left( y-{{y}_{1}} \right)}{\left( x-{{x}_{1}} \right)}$. The equation becomes $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$.
Note: We could have used the same process where we replace the origin value $O\left( 0,0 \right)$ with the point $Q\left( {{x}_{1}},{{y}_{1}} \right)$. The shifting of the origin will be sufficed to do that.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Trending doubts
State and prove Bernoullis theorem class 11 physics CBSE

Raindrops are spherical because of A Gravitational class 11 physics CBSE

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

Write the differences between monocot plants and dicot class 11 biology CBSE

Why is steel more elastic than rubber class 11 physics CBSE

Explain why a There is no atmosphere on the moon b class 11 physics CBSE
