Length and area - 5G
7. A point P lies on the line with equation y=4−3*x. The point P is a distance √(,34) from the origin. Find the two possible positions of point P.
P=((x_p),(y_p))
O=(0,0)
OP=√(,34)
=√(,((x_p)-0)^2+((y_p)-0)^2)
=√(,(x_p)^2+(y_p)^2)
∴(x_p)^2+(y_p)^2=34
P∈y=-3*x+4
∴ (y_p)=-3*(x_p)+4
P=((x_p),(y_p)):{[(y_p)=-3*(x_p)+4],[(x_p)^2+(y_p)^2=34])
(y_p)=-3*(x_p)+4⇒(x_p)^2+(y_p)^2=34
∴(x_p)^2+(-3*(x_p)+4)^2=34
(x_p)^2+(9*(x_p)^2-24*(x_p)+16)=34
*10*(x_p)^2-24*(x_p)-18=0
*5*(x_p)^2-12*(x_p)-9=0
5*(x_p)^2-15*(x_p)+3*(x_p)-9=0
(5*(x_p)+3)*((x_p)-3)=0
(x_p)={-3/5,3}
(y_p)=-3*(x_p)+4
(y_p)={29/5,-5}
∴P={(-3/5,29/5),(3,-5)}
8. The vertices of a triangle are A=(2,7), B=(5,−6) and C=(8,−6).
Show that the triangle is a scalene triangle.
ABC is scalene →AB≠BC≠AC
AB=√(,((x_A)-(x_B))^2+((y_A)-(y_B))^2)
=√(,(2-5)^2+(7+6)^2)
=√(,9+169)
=√(,178)
BC=√(,((x_B)-(x_C))^2+((y_B)-(y_C))^2)
=√(,(5-8)^2+(-6+6)^2)
=√(,9+0)
=3
AC=√(,((x_A)-(x_C))^2+((y_A)-(y_C))^2)
=√(,(2-8)^2+(7+6)^2)
=√(,36+169)
=√(,205)
√(,178)≠3≠√(,205)∴ABC is scalene.
Find the area of the triangle ABC.
(𝙰_)(ABC)=(b*h)/2=((8-5)*(7+6))/2=39/2
The straight line (l_1) has equation y=7*x−3. The straight line (l_2) has equation 4*x+3*y−41=0. The lines intersect at the point A.
Work out the coordinates of A.
A={[y=7*x−3],[4*x+3*y−41=0])
Sub y=7*x−3⇒4*x+3*y−41=0
4*(x_A)+3*(7*(x_A)−3)-41=0
-50+25*(x_A)=0
25*(x_A)=50
(x_A)=2
(y_A)=7*(2)-3
(y_A)=11
A=((x_A),(y_A))=(2,11)
The straight line (l_2) crosses the x-axis at the point B.
Work out the coordinates of B.
(l_2):4*(x_B)+3*(y_B)−41=0
4*(x_B)+3*(0)-41=0
4*(x_B)=41
(x_B)=41/4
4*(x_B)+3*(y_B)−41=0
4*(41/4)+3*(y_B)−41=0
B=(41/4,0)
Work out the area of triangle AOB.
(𝙰_)(AOB)=(𝚋*𝚑)/2=(41/4×11)/2=451/8
The straight line (l_1) has equation 4*x−5*y−10=0 has x-intercept at A.
The straight line (l_2) has equation 4*x−2*y+20=0 has x-intercept at B.
The straight lines (l_1) and (l_2) intersect at the point C.
Work out the coordinates of A, Band C.
A:4*(x_A)−5*(0)−10=0
(x_A)=5/2
A=(5/2,0)
B:4*(x_B)−2*(0)+20=0
(x_A)=5
B=(5,0)
C∈(l_1),(l_2)
C=((x_C),(y_C)):{[4*(x_C)−5*(y_C)−10=0],[4*(x_C)−2*(y_C)+20=0])
-3*(y_C)-30=0
3*(y_C)+30=0
(y_C)+10=0
(y_C)=-10
4*(x_C)-5*(-10)-10=0
(x_C)=-10
∴C=(-10,-10)
Work out the area of △ABC.
(𝙰_)(△ABC)=(𝚋*𝚑)/2=(5/2×10)/2=25/2
The points R=(5,−2) and S=(9,0) lie on the straight line (l_1) as shown.
The straight line (l_2) is perpendicular to (l_1) and passes through the point R.
Work out an equation for the straight lines (l_1) and (l_2).
𝓂((l_1))=(0--2)/(9-5)=2/4=1/2
(l_1):y=1/2*x+c
R∈(l_1)
R=(5,-2)
-2=1/2*(5)+c
c=-9/2
∴(l_1):1/2*x-9/2
𝓂((l_1))*𝓂((l_2))=-1
1/2*𝓂((l_2))=-1
𝓂((l_2))=-2
(l_2):y=-2*x+c
-2=-2*(5)+c
c=8
∴(l_2):2*x+8
Write down the coordinates of T.
T=((x_T),(y_T))
(x_T)=0
(y_T)=(c_(l_2))=8
∴T=(0,8)
Work out the lengths of RS and TR leaving your answer in the form k√(,5).
RS=√(,((x_R)-(x_S))^2+((y_R)-(y_S))^2)
=√(,(5-9)^2+(2-0)^2)
=√(,16+4)
=√(,20)
=2√(,5)
TR=√(,((x_T)-(x_R))^2+((y_T)-(y_R))^2)
=√(,(0-5)^2+(8+2)^2)
=√(,25+100)
=√(,125)
=5√(,5)
Work out the area of △RST.
(𝙰_)(△RST)=(𝚋*𝚑)/2=(RS*TR)/2=(2√(,5)×5√(,5))/2=25
The straight line (l_1) passes through the point (−4,14) and has gradient −1/4.
Find an equation for (l_1) in the form a*x+b*y+c=0, where a, b and c are integers.
(l_1):y=m*x+c
𝓂((l_1))=-1/4
∴(l_1):y=-1/4*x+c
(-4,14)∈(l_1)
→14=-1/4*(-4)+c
→14=1+c
→c=13
∴(l_1):y=-1/4*x+13
→1/4*x+y+13=0
→(l_1):x+4*y+52=0
Write down the coordinates of A, the point where straight line (l_1) crosses the y-axis.
(l_1):y=-1/4*x+13
c=13
A=(0,c)
A=(0,13)
The straight line (l_2) passes through the origin and has gradient 3. The lines (l_1) and (l_2) intersect at the point B.
Calculate the coordinates of B.
(l_2):y=m*x+c
𝓂((l_2))=3
(0,0)∈(l_2)
∴(l_2):y=3*x
B=((x_B),(y_B)):{[(y_B)=-1/4*(x_B)+13],[(y_B)=3*(x_)])
∴3*(x_B)=-1/4*(x_B)+13
→13/4*(x_B)=13
→(x_B)=4
→(y_)=3*(4)=12
∴B=((x_B),(y_B))=(4,12)
Calculate the exact area of △OAB.
(𝙰_)(AOB)=(𝚋*𝚑)/2=((y_A)*(x_B))/2=(13×4)/2=26