The equation of two equal sides AB and AC of an isosceles triangle ABC are x+y=5 &7x-y=3 respectively. Find the equations of the sides BC if the area of the triangle of ABC is 5 units
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a.)x+3y−1=0b.)x−3y+1=0c.)2x−y−5=0d.)x+2y−5=0a.)x+3y−1=0b.)x−3y+1=0c.)2x−y−5=0d.)x+2y−5=0

Let the slope of the required line be mm. I used the slope formula between two lines∠B=∠C ∣∣∣(−1−m)(1−m)∣∣∣=∣∣∣(7−m)(1+7m)∣∣∣⟹m=−3,13∠B=∠C |(−1−m)(1−m)|=|(7−m)(1+7m)|⟹m=−3,13
But the book is giving right answer as option d.)d.).
By finding mm i only found slope not the whole equation.
Also i would like to know if their is clean simple short way , and also using Area=5Area=5 .
I have studied maths upto 12th12th grade.
Update : by using Geogebra I found that x−3y+1=0x−3y+1=0 fits perfectly

Let the lines: x+y=5x+y=5 & 7x−y=37x−y=3represent AC & AB. Then the acute angle θθbetween the equal sides AB & AC is given as
tanθ=∣∣∣7−(−1)1+7(−1)∣∣∣=∣∣∣8−6∣∣∣=43⟹sinθ=45tanθ=|7−(−1)1+7(−1)|=|8−6|=43⟹sinθ=45
The length of two sides AB & AC of isosceles triangle are equal hence, the area of triangle is given as
12(AB)(AC)sinθ=5⟹12(AB)2(45)=5⟹AB=52–√12(AB)(AC)sinθ=5⟹12(AB)2(45)=5⟹AB=52
The equations of lines bisecting the angle between the equal sides: x+y=5x+y=5 & 7x−y=37x−y=3 are given as
x+y−512+12−−−−−−√=±7x−y−372+(−1)2−−−−−−−−−√⟹5x+5y−25=±(7x−y−3)x+y−512+12=±7x−y−372+(−1)2⟹5x+5y−25=±(7x−y−3)
⟹x−3y+11=0&3x+y−7=0⟹x−3y+11=0&3x+y−7=0
From above, equations it's clear that 3x+y−7=03x+y−7=0 is acute angle bisector having negative slope & passing through vertex A. We can also check it out by a rough sketch.
Let the equation of third (unknown) line i.e. side BC be x−3y+c=0x−3y+c=0 normal to the angle bisector: 3x+y−7=03x+y−7=0. Now, solving: x−3y+c=0x−3y+c=0 & any given line say 7x−y=37x−y=3, we get the intersection point (c+920,7c+320)(c+920,7c+320). Similarly, solving both the equations of the given lines, we get intersection point (1,4)(1,4). Now, calculating the length AB=52√AB=52 of equal sides of isosceles triangle using point-to-point distance-formula as
AB=(1−c+920)2+(4−7c+320)2−−−−−−−−−−−−−−−−−−−−−−−−−√=52–√AB=(1−c+920)2+(4−7c+320)2=52
(c−11)2=100⟹c−11=±10⟹c=21&c=1(c−11)2=100⟹c−11=±10⟹c=21&c=1
Thus by setting the values of cc, we get two equations of parallel lines representing the third unknown side BC as: x−3y+1=0x−3y+1=0 & x−3y+21=0x−3y+21=0lying on either side of vertex A of given isosceles triangle ABC. Thus, we see that the option (b) x−3y+1=0x−3y+1=0 is correct.
According to the option (d) in your book the unknown side is: x+2y−5=0x+2y−5=0. Note that this line has slope −12−12 i.e. negative but see in the figure you drew the slope of unknown side must be only positive. Thus option provided in your book is absolutely wrong (may be due to some printing mistake).

Let the slope of the required line be mm. I used the slope formula between two lines∠B=∠C ∣∣∣(−1−m)(1−m)∣∣∣=∣∣∣(7−m)(1+7m)∣∣∣⟹m=−3,13∠B=∠C |(−1−m)(1−m)|=|(7−m)(1+7m)|⟹m=−3,13
But the book is giving right answer as option d.)d.).
By finding mm i only found slope not the whole equation.
Also i would like to know if their is clean simple short way , and also using Area=5Area=5 .
I have studied maths upto 12th12th grade.
Update : by using Geogebra I found that x−3y+1=0x−3y+1=0 fits perfectly

Let the lines: x+y=5x+y=5 & 7x−y=37x−y=3represent AC & AB. Then the acute angle θθbetween the equal sides AB & AC is given as
tanθ=∣∣∣7−(−1)1+7(−1)∣∣∣=∣∣∣8−6∣∣∣=43⟹sinθ=45tanθ=|7−(−1)1+7(−1)|=|8−6|=43⟹sinθ=45
The length of two sides AB & AC of isosceles triangle are equal hence, the area of triangle is given as
12(AB)(AC)sinθ=5⟹12(AB)2(45)=5⟹AB=52–√12(AB)(AC)sinθ=5⟹12(AB)2(45)=5⟹AB=52
The equations of lines bisecting the angle between the equal sides: x+y=5x+y=5 & 7x−y=37x−y=3 are given as
x+y−512+12−−−−−−√=±7x−y−372+(−1)2−−−−−−−−−√⟹5x+5y−25=±(7x−y−3)x+y−512+12=±7x−y−372+(−1)2⟹5x+5y−25=±(7x−y−3)
⟹x−3y+11=0&3x+y−7=0⟹x−3y+11=0&3x+y−7=0
From above, equations it's clear that 3x+y−7=03x+y−7=0 is acute angle bisector having negative slope & passing through vertex A. We can also check it out by a rough sketch.
Let the equation of third (unknown) line i.e. side BC be x−3y+c=0x−3y+c=0 normal to the angle bisector: 3x+y−7=03x+y−7=0. Now, solving: x−3y+c=0x−3y+c=0 & any given line say 7x−y=37x−y=3, we get the intersection point (c+920,7c+320)(c+920,7c+320). Similarly, solving both the equations of the given lines, we get intersection point (1,4)(1,4). Now, calculating the length AB=52√AB=52 of equal sides of isosceles triangle using point-to-point distance-formula as
AB=(1−c+920)2+(4−7c+320)2−−−−−−−−−−−−−−−−−−−−−−−−−√=52–√AB=(1−c+920)2+(4−7c+320)2=52
(c−11)2=100⟹c−11=±10⟹c=21&c=1(c−11)2=100⟹c−11=±10⟹c=21&c=1
Thus by setting the values of cc, we get two equations of parallel lines representing the third unknown side BC as: x−3y+1=0x−3y+1=0 & x−3y+21=0x−3y+21=0lying on either side of vertex A of given isosceles triangle ABC. Thus, we see that the option (b) x−3y+1=0x−3y+1=0 is correct.
According to the option (d) in your book the unknown side is: x+2y−5=0x+2y−5=0. Note that this line has slope −12−12 i.e. negative but see in the figure you drew the slope of unknown side must be only positive. Thus option provided in your book is absolutely wrong (may be due to some printing mistake).
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