The points (-6,1), (6,10), (9,6) and (-3,-3) are the vertices of a rectangle If the area of the portion - Maths - Straight Lines

Question

The points (-6,1), (6,10), (9,6) and (-3,-3) are the vertices of a
rectangle If the area of the portion of this rectangle lies above
x-axis is a+b where (a,b C N) and a and b are in their lowest form
Then find a+b

Anuradha Sharma · Accepted Answer

Dear Student,
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"https://s3mn%20mnimgs%20com/img/shared/content_ck_images/ana_qa_image_57b2f09599329%20png">
Equation of line AD is , y+3 = 4-3x+3Now if it cuts x-axis then y = 0 4x+12= -9 x = -214=-5
25So coordinate of F = -5
25,0Similarly equation of line AB is , y+3 = 912x+3Now if it cuts x-axis then y =012= 3x+9 x = 1 so coordinate of E is 1,0Now area of triange AEF using herons formula,  =12×EF×3=32×6
25=9
375 square unitNow find area of square ABCD by finding side AB and Bc and then multiply them anand subtract area of AEF from it and then that will be a+b
 For simplest form represent that area of a+b in terms of sum of two prime numbers
 Regards

The points (-6,1), (6,10), (9,6) and (-3,-3) are the vertices of a rectangle. If the area of the portion of this rectangle lies above x-axis is a+b where (a,b C N) and a and b are in their lowest form. Then find a+b