A line passes through the points (1, 4) and (5, 8). A second line passes through the points (2, 10) and (6, 4). At what point do the two lines intersect?
Answers
Answer:
Slope(m)(m) of line passing through (x_1,y_1)\ and\ (x_2,y_2)(x
1
,y
1
) and (x
2
,y
2
) is given by \frac{y_2-y_1}{x_2-x_1}
x
2
−x
1
y
2
−y
1
.
Slope intercept form of line is given by: y=mx+by=mx+b
First Line:
Let first line is y=m_1x+b_1y=m
1
x+b
1
\begin{gathered}m_1=\frac{8-4}{5-1}=\frac{4}{4}=1\\\\Line:\ y=x+b_1\\\\It\ passes\ through\ (1,4)\\\\4=1+b_1\Rightarrow\ b_1=3\\\\First\ line\ y=x+3\end{gathered}
m
1
=
5−1
8−4
=
4
4
=1
Line: y=x+b
1
It passes through (1,4)
4=1+b
1
⇒ b
1
=3
First line y=x+3
Second Line:
Let second line is y=m_2x+b_2y=m
2
x+b
2
\begin{gathered}m_2=\frac{4-10}{6-2}=\frac{-6}{4}=-\frac{3}{2}\\\\Line:\ y=-\frac{3}{2}x+b_2\\\\It\ passes\ through\ (2,10)\\\\10=-3+b_2\Rightarrow\ b_2=13\\\\Second\ line\ y=-\frac{3}{2}x+13\end{gathered}
m
2
=
6−2
4−10
=
4
−6
=−
2
3
Line: y=−
2
3
x+b
2
It passes through (2,10)
10=−3+b
2
⇒ b
2
=13
Second line y=−
2
3
x+13
Point of Intersection of Two Lines:
\begin{gathered}Solve\\\\y=x+3\\\\y=-\frac{3}{2}x+13\\\\\\x+3=-\frac{3}{2}x+13\\\\x+\frac{3}{2}=13-3\\\\\frac{5}{2}x=10\\\\x=\frac{2}{5}\times 10\\\\x=4\\\\Since\ y=x+3\\\\y=4+3=7\\\\Point\ of\ intersection\ is\ (4,7)\end{gathered}
Solve
y=x+3
y=−
2
3
x+13
x+3=−
2
3
x+13
x+
2
3
=13−3
2
5
x=10
x=
5
2
×10
x=4
Since y=x+3
y=4+3=7
Point of intersection is (4,7)
Step-by-step explanation:
two line intersect at 4,7.
this is the answer.
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