What is the solution to the system of liner equations shown on the graph?
1 answer:
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First, write the equation of the line containing the points <span>(2,-5) and (-3,2).
We can use 2 point form, or point-slope form.
Let's use </span>point-slope form.
the slope m is
, then use any of the points to write the equation. (ex, pick (2, -5))
y-(-5)=(-7/5)(x-2)
y+5=(-7/5)x+14/5
y= (-7/5)x+14/5 - 5 =(-7/5)x+14/5 - 25/5 =(-7/5)x-11/5
Thus, the lines are
i) y=-ax+4 and ii) y=(-7/5)x-11/5
the slopes are the coefficients of x: -a and (-7/5),
the product of the slopes of 2 perpendicular lines is -1,
so
(-a)(-7/5)=-1
7/5a=-1
a=-1/(7/5)=-5/7
Answer: -5/7
We can use 2 point form, or point-slope form.
Let's use </span>point-slope form.
the slope m is
y-(-5)=(-7/5)(x-2)
y+5=(-7/5)x+14/5
y= (-7/5)x+14/5 - 5 =(-7/5)x+14/5 - 25/5 =(-7/5)x-11/5
Thus, the lines are
i) y=-ax+4 and ii) y=(-7/5)x-11/5
the slopes are the coefficients of x: -a and (-7/5),
the product of the slopes of 2 perpendicular lines is -1,
so
(-a)(-7/5)=-1
7/5a=-1
a=-1/(7/5)=-5/7
Answer: -5/7