Question

Problem 4: (a) (5 points) Find the equation for the line that passes through the points (-4,-2) and (8, 1). Write your equation in se form, slope-intercept form, or point- slope form. (Extra Credit: Write the equation for the line in all three forms) (b) (5 points) Graph the line. Problem 5: (10 points) Find the equation for the line passing through the point (3.2) and perpendicular to the line y = ',x + 7. The the line Problem 6: (10 points) Solve the quadratic equation x2 -7x+12=0. Use any method you would like. Problem 7: (10 points) Solve the quadratic equation 3x? -5x-2=0. Use any method you would like.

Answer 1

Q4. a)

& ( 81) xiyi y Slope Formula m= Y, - y 341 - 2 m= 3 12 8-(-4) ។ Slop Interent form of equation y = matb y = 4 ។ To find b, we will put the values (-4,-2) in equation 5 - 2= 1.(-4+ b 4 & Slope Interecht Form is y = Loc-| 4 чух - Ч. Standard Form of equation is I-4 4 2C-4 y = 4 Point Slope Form is - ( (y- y) = m(x-3) (y - 1) = 1 Gug) 4 4

b)

Yarar 2 543 -1 ० । 2 7 १ хшая -2 (-4-2) -3

Q5.

Sloke of perfontuber line: als o P (3,2) y of Pxt7 4 m. m=slope of line L ។ -1 -4 Equation Form) y = mx + b y = -4 oct b Putting the values (3,2) to find b (Interecht) 2 = -463) + b b = 2 +12 b = 14 & Equation of the line is => -4x+14 성트 22

Q6.

Dc2-7x+12=0 D = 62-4ac D=670²_400(12) 49-48 = 1 Ic= -b + ſo 2 a Sc= 7 £ v = 2 = 7+1=y 2 2 = - 7-1 2 C = 4,3

Q7.

3x² - 50c-2=0 2 S D= 6²-Yac D = 653 – 4(3)(-2) = 25+24=49 + C = -btro - (5) tryg 2 (3) 2 а 2 5 +7 6 I= 12 = 2 6 = = -2 6 3