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Question
1)For what value of k will the graph of 6x + ky = 6 be perpendiculuar to the graph of 2x – 6y = 12?
2)Write the standard form of the equation of the line parallel to the graph of x – 2y – 6 = 0 and passing through C(0, 1).
3)Write an equation of the line perpendicular to the graph of x = 3 and passing through D(4, –1).
Hi Bob,
The product of slopes of two lines must be - 1.
6x + ky = 6,
ky = -6x + 6
y = - (6/k)x + (6/k) ..... ( 1 )
2x - 6y = 12
-6y = -2x + 12
6y = 2x - 12
y = (2/6)x - 2 ....... ( 2 )
{(-6/k)}(2/6)= -1
k = 2
You may write to me at sc.benjamin@yahoo.com for regular online interaction.
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Suresh Chandra Benjamin
Expertise
I would like to answer middle school Algebra,Geometry and General Math questions and love to answer WORD PROBLEMs.
Experience
I started teaching Maths and Physics to the students of grades 9 and 10, soon after completing postgraduation in Physics ,in a renownned English medium school in north India in 1975.
Organizations
I have served as a Physics teacher in a renowned English medium school in north India for more than two decades and as a Principal of a K-12 school for four years.Currently I am nurturing Benjamin Academia - an Institute to help 6 to 12 graders.
Publications
A text book in two volumes for grades 9 and 10 :
" A Complete Course in ICSE Physics " published by S.Chand and Company,Delhi.
Education/Credentials
I have a Bachelor's degree in Math,Physics and Chemistry and Master's degree in Physics.
Awards and Honors
I was awarded "Crusade Scholarship" by Methodist Church in India.
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